tipler mosca physics for scientists and engineers solutions

Chapter 10

742

Find the moment of inertia Ia of the sun when it has collapsed into a spherical neutron star of radius 10 km and uniform mass distribution:

( )( )237

23052

252

a

mkg1096.7

km10kg1099.1

⋅×=

×=

= MRI

Substitute in equation (1) and solve for ωa to obtain:

b8

b237

246

ba

ba

1015.7

mkg1096.7mkg1069.5

ω

ωωω

×=

⋅×⋅×

==II

Given that ωb = 1 rev/25 d, evaluate ωa:

rev/d1086.2

d25rev11015.7

7

8a

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛×=ω

smaller. getssun theas decreases which energy, potential nalgravitatioof expense at the comesenergy kinetic rotational additional The

Note that the rotational period decreases by the same factor of Ib/Ia and becomes:

s1002.3

s3600h1

h24d1

revrad2

drev1086.2

22 3

7aa

−×=××××

== πT πωπ

(b) Express the fractional change in the sun’s rotational kinetic energy as a consequence of its collapse and simplify to obtain:

1

1

1

2bb

2aa

2bb2

1

2aa2

1

b

a

b

ba

b

−=

−=

−=−

=∆

ωω

ωω

II

II

KK

KKK

KK

Substitute numerical values and evaluate ∆K/Kb:

827

8b

1015.71drev/251

rev/d102.861015.7

1×=−⎟⎟

⎠

⎞⎜⎜⎝

⎛ ×⎟⎠⎞

⎜⎝⎛

×=

∆KK

(i.e., the rotational kinetic

energy increases by a factor of approximately 7×108.) 29 •• Picture the Problem We can solve 2CMRI = for C and substitute numerical values in order to determine an experimental value of C for the earth. We can then compare this value to those for a spherical shell and a sphere in which the mass is uniformly distributed to decide whether the earth’s mass density is greatest near its core or near its crust.

Conservation of Angular Momentum

743

(a) Express the moment of inertia of the earth in terms of the constant C:

2CMRI =

Solve for C to obtain: 2MR

IC =

Substitute numerical values and evaluate C: ( )( )

331.0

km6370kg105.98mkg108.03

224

237

=

×⋅×

=C

(b) If all of the mass were in the crust, the moment of inertia of the earth would be that of a thin spherical shell:

232

shell spherical MRI =

If the mass of the earth were uniformly distributed throughout its volume, its moment of inertia would be:

252

sphere solid MRI =

earth. theofcenter near thegreater bemust density mass the0.4, 2/5 ally experiment Because =<C

*30 •• Picture the Problem Let’s estimate that the diver with arms extended over head is about 2.5 m long and has a mass M = 80 kg. We’ll also assume that it is reasonable to model the diver as a uniform stick rotating about its center of mass. From the photo, it appears that he sprang about 3 m in the air, and that the diving board was about 3 m high. We can use these assumptions and estimated quantities, together with their definitions, to estimate ω and L. Express the diver’s angular velocity ω and angular momentum L:

t∆∆

=θω (1)

and ωIL = (2)

Using a constant-acceleration equation, express his time in the air:

gy

gy

ttt

downup

m 6 fallm 3 rise

22 ∆+

∆=

∆+∆=∆

Substitute numerical values and evaluate ∆t:

( ) ( ) s89.1m/s9.81m62

m/s9.81m32

22 =+=∆t

Estimate the angle through which he rotated in 1.89 s:

radrev5.0 πθ =≈∆

Chapter 10

744

Substitute in equation (1) and evaluate ω: rad/s66.1

s89.1rad

==πω

Use the ″stick rotating about an axis through its center of mass″ model to approximate the moment of inertia of the diver:

2121 MLI =

Substitute in equation (2) to obtain:

ω2121 MLL =

Substitute numerical values and evaluate L:

( )( ) ( )/smkg70/smkg2.69

rad/s1.66m2.5kg8022

2121

⋅≈⋅=

=L

Remarks: We can check the reasonableness of this estimation in another way. Because he rose about 3 m in the air, the initial impulse acting on him must be about 600 kg⋅m/s (i.e., I = ∆p = Mvi). If we estimate that the lever arm of the force is roughly l = 1.5 m, and the angle between the force exerted by the board and a line running from his feet to the center of mass is about 5°, we obtain °= sin5IL l ≈ 78 kg⋅m2/s, which is not too bad considering the approximations made here. 31 •• Picture the Problem First we assume a spherical diver whose mass M = 80 kg and whose diameter, when curled into a ball, is 1 m. We can estimate his angular velocity when he has curled himself into a ball from the ratio of his angular momentum to his moment of inertia. To estimate his angular momentum, we’ll guess that the lever arml of the force that launches him from the diving board is about 1.5 m and that the angle between the force exerted by the board and a line running from his feet to the center of mass is about 5°. Express the diver’s angular velocity ω when he curls himself into a ball in mid-dive:

IL

=ω (1)

Using a constant-acceleration equation, relate the speed with which he left the diving board v0 to his maximum height ∆y and our estimate of his angle with the vertical direction:

yav yy ∆+= 20 20

where °= 5cos00 vv y

Solve for v0:

°∆

=5cos

220

ygv

Substitute numerical values and evaluate v0:

( )( )m/s7.70

5cosm3m/s9.812 2

0 =°

=v


745

Approximate the impulse acting on the diver to launch him with the speed v0:

0MvpI =∆=

Letting l represent the lever arm of the force acting on the diver as he leaves the diving board, express his angular momentum:

°=°= 5sin5sin 0ll MvIL

Use the ″uniform sphere″ model to approximate the moment of inertia of the diver:

252 MRI =

Substitute in equation (1) to obtain: 2

02

520

25sin55sin

Rv

MRMv °

=°

=llω

Substitute numerical values and evaluate ω:

( )( )( )

rad/s10.1

m0.525sinm1.5m/s7.705

2

=

°=ω

*32 •• Picture the Problem We’ll assume that he launches himself at an angle of 45° with the horizontal with his arms spread wide, and then pulls them in to increase his rotational speed during the jump. We’ll also assume that we can model him as a 2-m long cylinder with an average radius of 0.15 m and a mass of 60 kg. We can then find his take-off speed and ″air time″ using constant-acceleration equations, and use the latter, together with the definition of rotational velocity, to find his initial rotational velocity. Finally, we can apply conservation of angular momentum to find his initial angular momentum. Using a constant-acceleration equation, relate his takeoff speed v0 to his maximum elevation ∆y:

yavv yy ∆+= 220

2

or, because v0y = v0sin45°, v = 0, and ay = − g,

ygv ∆−°= 245sin0 220

Solve for v0 to obtain:

°∆

=°

∆=

45sin2

45sin2

20ygygv


( )( )m/s4.85

sin45m0.6m/s9.812 2

0 =°

=v

Use its definition to express Goebel’s angular velocity: t∆

∆=

θω

Use a constant-acceleration equation to express Goebel’s ″air time″ ∆t: g

ytt ∆=∆=∆

222 m 0.6 rise

Chapter 10

746


( ) s699.0m/s9.81

m6.022 2 ==∆t

Substitute numerical values and evaluate ω: rad/s0.36

revrad2π

s0.699rev4

=×=ω

Use conservation of angular momentum to relate his take-off angular velocity ω0 to his average angular velocity ω as he performs a quadruple Lutz:

ωω II =00

Assuming that he can change his angular momentum by a factor of 2 by pulling his arms in, solve for and evaluate ω0:

( ) rad/s18.0rad/s3621

00 === ωω

II

Express his take-off angular momentum:

000 ωIL =

Assuming that we can model him as a solid cylinder of length l with an average radius r and mass m, express his moment of inertia with arms drawn in (his take-off configuration):

( ) 2221

0 2 mrmrI == where the factor of 2 represents our assumption that he can double his moment of inertia by extending his arms.

Substitute to obtain: 0

20 ωmrL =

Substitute numerical values and evaluate L0:

( )( ) ( )/smkg3.24

rad/s18m0.15kg602

20

⋅=

=L

Vector Nature of Rotation 33 • Picture the Problem We can express F

rand r

rin terms of the unit vectors i and j and

then use the definition of the cross product to find .τr Express F

rin terms of F and the unit

vector :i

iF ˆF−=r

Express rr

in terms of R and the unit vector :j

jr ˆR=r


747

Calculate the cross product of rr

and :F

r

( )( ) kji

ijFrτˆˆˆ

ˆˆ

FRFR

FR

=×=

−×=×=rrr

34 • Picture the Problem We can find the torque is the cross product of r

rand .F

r

Compute the cross product of rr and :F

r ( )( )

( ) ( )k

jjji

jjiFrτ

ˆ

ˆˆˆˆ

ˆˆˆ

mgx

mgymgx

mgyx

−=

×−×−=

−+=×=rrr

35 • Picture the Problem The cross product of the vectors jiA ˆˆ

yx AA +=r

and jiB ˆˆyx BB +=

ris given by

( ) ( ) ( ) ( )jjijjiiiBA ˆˆˆˆˆˆˆˆ ×+×+×+×=× yyxyyxxx BABABABArr

( ) ( ) ( ) ( )0ˆˆ0 yyxyyxxx BABABABA +−++= kk

( ) ( )kk ˆˆ −+= xyyx BABA

(a) Find A

r× Br

for Ar

= 4 i and Br

= 6 i + j6 : ( )

( ) ( )( ) kk

jiii

jiiBA

ˆ24ˆ24024

ˆˆ24ˆˆ24

ˆ6ˆ6ˆ4

=+=

×+×=

+×=×rr

(b) Find A

r× Br

for Ar

= 4 i and Br

= 6 i + 6 k : ( )

( ) ( )( ) ( ) jj

kiii

kiiBA

ˆ24ˆ24024

ˆˆ24ˆˆ24

ˆ6ˆ6ˆ4

−=−+=

×+×=

+×=×rr

(c) Find Ar

× Br

for Ar

= 2 i + j3

and Br

=3 i + j2 :

( ) ( )( ) ( ) ( )

( )( ) ( ) ( ) ( )

k

kk

jj

ijjiii

jijiBA

ˆ5

06ˆ9ˆ406

ˆˆ6

ˆˆ9ˆˆ4ˆˆ6

ˆ2ˆ3ˆ3ˆ2

−=

+−++=

×+

×+×+×=

+×+=×rr

Chapter 10

748

*36 • Picture the Problem The magnitude of A

r× Br

is given by θsinAB .

Equate the magnitudes of A

r× Br

and BA

rr⋅ :

θθ cossin ABAB =

θθ cossin =∴

or 1tan ±=θ

Solve for θ to obtain: °±°±=±= − 135or451tan 1θ

37 •• Picture the Problem Let rr be in the xy plane. Then ωr points in the positive z direction. We can establish the results called for in this problem by forming the appropriate cross products and by differentiating .vr

(a) Express ωr using unit vectors: kω ˆω=r

Express r

rusing unit vectors: ir ˆr=

r

Form the cross product of ω

rand :rr ( )

j

jikikrωˆ

ˆˆˆˆˆ

v

rrr

=

=×=×=× ωωωrr

rωv

rrr×=∴

(b) Differentiate v

rwith respect to t to

express ar

: ( )

( )ct

t

aarωωa

vωrω

rωrω

rωva

rr

rrrr

rrrr

rrr

r

rrr

r

+=××+=

×+×=

×+×=

×==

dtd

dtd

dtd

dtd

dtd

where ( )rωωa rrrr××=c

and ct and aa rrare the tangential and


749

centripetal accelerations, respectively.

38 •• Picture the Problem Because Bz = 0, we can express B

ras jiB ˆˆ

yx BB +=r

and form its

cross product with Ar

to determine Bx and By. Express B

rin terms of its components: jiB ˆˆ

yx BB +=r

(1)

Express A

r× Br

: ( ) kkjiiBA ˆ12ˆ4ˆˆˆ4 ==+×=× yyx BBBrr

Solve for By: 3=yB

Relate B to Bx and By: 222

yx BBB +=

Solve for and evaluate Bx: 435 2222 =−=−= yx BBB

Substitute in equation (1): jiB ˆ3ˆ4 +=

r

39 • Picture the Problem We can write B

rin the form kjiB ˆˆˆ

zyx BBB ++=r

and use the dot

product of Ar

and Br

to find By and their cross product to find Bx and Bz.

Express Br

in terms of its components: kjiB ˆˆˆzyx BBB ++=

r (1)

Evaluate A

r⋅ :Br

123 ==⋅ yBBArr

and By = 4

Evaluate Ar

× :Br

( )ik

kjijBAˆ3ˆ3

ˆˆ4ˆˆ3

zx

zx

BB

BB

+−=

++×=×rr

Because A

r× Br

= 9 :i Bx = 0 and Bz = 3.

Substitute in equation (1) to obtain: kjB ˆ3ˆ4 +=r

Chapter 10

750

40 •• Picture the Problem The dot product of A

rwith the cross product of B

rand C

ris a scalar

quantity and can be expressed in determinant form as

zyx

zyx

zyx

cccbbbaaa

. We can expand this

determinant by minors to show that it is equivalent to )( CBArrr

×⋅ , )( BACrrr

×⋅ , and

)( ACBrrr

×⋅ . The dot product of A

rwith the cross

product of Br

and Cr

is a scalar quantity and can be expressed in determinant form as:

zyx

zyx

zyx

cccbbbaaa

=×⋅ )( CBArrr

Expand the determinant by minors to obtain:

xyzyxz

zxyxzy

yzxzyx

zyx

zyx

zyx

cbacba

cbacba

cbacbacccbbbaaa

−+

−+

−=

(1)

Evaluate the cross product of B

rand

Cr

to obtain:

( )( ) ( )kj

iCBˆˆ

ˆ

xyyxzxxz

yzzy

cbcbcbcb

cbcb

−+−+

−=×rr

Form the dot product of A

rwith

Br

× Cr

to obtain:

( )

xyzyxz

zxyxzy

yzxzyx

cbacba

cbacba

cbacba

−+

−+

−=×⋅ CBArrr

(2)

Because (1) and (2) are the same, we can conclude that:

zyx

zyx

zyx

cccbbbaaa

=×⋅ )( CBArrr

Proceed as above to establish that:

zyx

zyx

zyx

cccbbbaaa

=×⋅ )( BACrrr

and


751

zyx

zyx

zyx

cccbbbaaa

=×⋅ )( ACBrrr

41 •• Picture the Problem Let, without loss of generality, the vector C

rlie along the x axis and

the vector Br

lie in the xy plane as shown below to the left. The diagram to the right shows the parallelepiped spanned by the three vectors. We can apply the definitions of the cross- and dot-products to show that ( )CBA

rrr×⋅ is the volume of the parallelepiped.

Express the cross-product of B

rand :C

r ( )( )kCB ˆsin −=× θBC

rr

and ( )

ramparallelog theof area

sin

=

=× CB θCBrr

Form the dot-product of A

rwith the

cross-product of Br

and Cr

to obtain: ( ) ( )

( )( )( )( )

ipedparallelep

heightbase of areacossin

cossin

V

ABCCBA

=

===×⋅

φθφθCBA

rrr

*42 •• Picture the Problem Draw the triangle using the three vectors as shown below. Note that .CBA

rrr=+ We can find the

magnitude of the cross product of Ar

and Br

and of Ar

and Cr

and then use the cross product of A

rand ,C

r using ,CBA

rrr=+ to

show that cABbAC sinsin = or B/sin b = C/ sin c. Proceeding similarly, we can extend the law of sines to the third side of the triangle and the angle opposite it.

Chapter 10

752

Express the magnitude of the cross product of A

rand :B

r

cABsin=× BArr

Express the magnitude of the cross product of A

rand :C

r

bAC sin=×CArr

Form the cross product of Ar

with Cr

to obtain:

( )

BA

BAAA

BAACA

rr

rrrr

rrrrr

×=

×+×=

+×=×

because 0=× AArr

.

Because :BACArrrr

×=× BACArrrr

×=×

and cABbAC sinsin =

Simplify and rewrite this expression to obtain: c

Cb

Bsinsin

=

Proceed similarly to extend this result to the law of sines: c

Cb

Ba

Asinsinsin

==

Angular Momentum 43 • Picture the Problem L

rand p

r are related according to .prL

rrr×= If L

r= 0, then

examination of the magnitude of pr rr× will allow us to conclude that 0sin =φ and that

the particle is moving either directly toward the point, directly away from the point, or through the point.

Because L

r= 0: 0=×=×=× vrvrpr

rrrrrrmm

or 0=× vr rr

Express the magnitude of :vr rr× 0sin ==× φrvvr rr

Because neither r nor v is zero: 0sin =φ

where φ is the angle between rr

and .vr

Solve for φ: °°== − 180or00sin 1φ


753

44 • Picture the Problem We can use their definitions to calculate the angular momentum and moment of inertia of the particle and the relationship between L, I, and ω to determine its angular speed.

(a) Express and evaluate the magnitude of :L

r

( )( )( )/smkg28.0

m4m/s3.5kg22⋅=

== mvrL

(b) Express the moment of inertia of the particle with respect to an axis through the center of the circle in which it is moving:

( )( ) 222 mkg32m4kg2 ⋅=== mrI

(c) Relate the angular speed of the particle to its angular momentum and solve for and evaluate ω:

22

2

rad/s0.875mkg32

/smkg28.0=

⋅⋅

==ILω

45 • Picture the Problem We can use the definition of angular momentum to calculate the angular momentum of this particle and the relationship between its angular momentum and angular speed to describe the variation in its angular speed with time.

(a) Express the angular momentum of the particle as a function of its mass, speed, and distance of its path from the reference point:

( )( )( )/smkg54.0

sin90m/s4.5kg2m6sin

2⋅=

°== θrmvL

(b) Because L = mr2ω: 1

2r∝ω and

recedes.it asdecreases andpoint theapproaches

particle theas increases ω

*46 •• Picture the Problem We can use the formula for the area of a triangle to find the area swept out at t = t1, add this area to the area swept out in time dt, and then differentiate this expression with respect to time to obtain the given expression for dA/dt. Express the area swept out at t = t1: 12

1112

11 cos bxbrA == θ

where θl is the angle between 1rr

and vr

and

Chapter 10

754

x1 is the component of 1rr

in the direction of vr .

Express the area swept out at t = t1 + dt:

( )( )vdtxb

dxxbdAAA+=

+=+=

121

121

1

Differentiate with respect to t: constant2

121 === bv

dtdxb

dtdA

Because rsinθ = b: ( ) ( )

mL

rpm

vrbv

2

sin21sin2

121

=

== θθ

47 •• Picture the Problem We can find the total angular momentum of the coin from the sum of its spin and orbital angular momenta. (a) Express the spin angular momentum of the coin:

spincmspin ωIL =

From Problem 9-44: 241 MRI =

Substitute for I to obtain: spin

241

spin ωMRL =

Substitute numerical values and evaluate Lspin:

( )( )

/smkg1033.1

revrad2

srev10

m0.0075kg0.015

25

241

spin

⋅×=

⎟⎠⎞

⎜⎝⎛ ××

=

−

π

L

(b) Express and evaluate the total angular momentum of the coin: /smkg1033.1

025

spinspinorbit

⋅×=

+=+=−

LLLL

(c) From Problem 10-14: 0orbit =L

and /smkg1033.1 25 ⋅×= −L

(d) Express the total angular momentum of the coin:

spinorbit LLL +=


755

Find the orbital momentum of the coin: ( )( )( )

/smkg107.50m0.1m/s0.05kg0.015

25

orbit

⋅×±=

±=±=

−

MvRL

where the ± is a consequence of the fact that the coin’s direction is not specified.

Substitute to obtain:

/smkg1033.1/smkg1050.7

25

25

⋅×+

⋅×±=−

−L

The possible values for L are: /smkg1083.8 25 ⋅×= −L

or /smkg1017.6 25 ⋅×−= −L

48 •• Picture the Problem Both the forces acting on the particles exert torques with respect to an axis perpendicular to the page and through point O and the net torque about this axis is their vector sum.

Express the net torque about an axis perpendicular to the page and through point O: ( ) 121

2211i

inet

Frr

FrFrττrrr

rrrrrr

×−=

×+×== ∑

because 12 FFrr

−=

Because 21 rr rr

− points along 1Fr

− : ( ) 0121 =×− Frrrrr

Torque and Angular Momentum 49 • Picture the Problem The angular momentum of the particle changes because a net torque acts on it. Because we know how the angular momentum depends on time, we can find the net torque acting on the particle by differentiating its angular momentum. We can use a constant-acceleration equation and Newton’s 2nd law to relate the angular speed of the particle to its angular acceleration.

(a) Relate the magnitude of the torque acting on the particle to the rate at which its angular momentum changes:

( )[ ]

mN00.4

mN4net

⋅=

⋅== tdtd

dtdLτ

Chapter 10

756

(b) Using a constant-acceleration equation, relate the angular speed of the particle to its acceleration and time-in-motion:

tαωω += 0

where ω0 = 0

Use Newton’s 2nd law to relate the angular acceleration of the particle to the net torque acting on it:

2netnet

mrIττα ==

Substitute to obtain: tmr 2

netτω =


( )( )( )( ) rad/s0.192

m3.4kg8.1mN4

2

2

t

t

=

⋅=ω

provided t is in seconds. 50 •• Picture the Problem The angular momentum of the cylinder changes because a net torque acts on it. We can find the angular momentum at t = 25 s from its definition and the net torque acting on the cylinder from the rate at which the angular momentum is changing. The magnitude of the frictional force acting on the rim can be found using the definition of torque. (a) Use its definition to express the angular momentum of the cylinder:

ωω 221 mrIL ==


( )( )

/smkg377

s60min1

revrad2

minrev500

m0.4kg90

2

221

⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛×××

=

π

L

(b) Express and evaluate dtdL

: ( )

22

2

/smkg15.1

s25/smkg377

⋅=

⋅=

dtdL

(c) Because the torque acting on the uniform cylinder is constant, the rate

22/smkg15.1 ⋅==dtdLτ


757

of change of the angular momentum is constant and hence the instantaneous rate of change of the angular momentum at any instant is equal to the average rate of change over the time during which the torque acts: (d) Using the definition of torque that relates the applied force to its lever arm, express the magnitude of the frictional force f acting on the rim:

N37.7m0.4

/smkg15.1 22

=⋅

==l

τf

*51 •• Picture the Problem Let the system include the pulley, string, and the blocks and assume that the mass of the string is negligible. The angular momentum of this system changes because a net torque acts on it.

(a) Express the net torque about the center of mass of the pulley: ( )12

12net

sin

sin

mmRg

gRmgRm

−=

−=

θ

θτ

where we have taken clockwise to be positive to be consistent with a positive upward velocity of the block whose mass is m1 as indicated in the figure.

(b) Express the total angular momentum of the system about an axis through the center of the pulley:

⎟⎠⎞

⎜⎝⎛ ++=

++=

212

21

mmRIvR

vRmvRmIL ω

(c) Express τ as the time derivative of the angular momentum:

⎟⎠⎞

⎜⎝⎛ ++=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++==

212

212

mmRIaR

mmRIvR

dtd

dtdLτ

Equate this result to that of part (a) and solve for a to obtain:

( )

212

12 sin

mmRI

mmga++

−=

θ

Chapter 10

758

52 •• Picture the Problem The forces resulting from the release of gas from the jets will exert a torque on the spaceship that will slow and eventually stop its rotation. We can relate this net torque to the angular momentum of the spaceship and to the time the jets must fire.

Relate the firing time of the jets to the desired change in angular momentum:

netnet τω

τ∆

=∆

=∆ILt

Express the magnitude of the net torque exerted by the jets:

FR2net =τ

Letting ∆m/∆t′ represent the mass of gas per unit time exhausted from the jets, relate the force exerted by the gas on the spaceship to the rate at which the gas escapes:

vtmF'∆

∆=

Substitute and solve for ∆t to obtain:

vRtm

It

'2

∆∆∆

=∆ω


( )( )( )( ) s52.4

m3m/s800kg/s102s60

min1rev

rad2minrev6mkg4000

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛××⋅

=∆ −

π

t

53 •• Picture the Problem We can use constant-acceleration equations to express the projectile’s position and velocity coordinates as functions of time. We can use these coordinates to express the particle’s position and velocity vectors r

rand .vr Using its

definition, we can express the projectile’s angular momentum Lr

as a function of time and then differentiate this expression to obtain .dtdL

r Finally, we can use the definition of

the torque, relative to an origin located at the launch position, the gravitational force exerts on the projectile to express τr and complete the demonstration that .τL rr

=dtd Using its definition, express the angular momentum vector L

r of the

projectile:

vrL rrrm×= (1)

Using constant-acceleration ( )tVtvx x θcos0 ==


759

equations, express the position coordinates of the projectile as a function of time:

and

( ) 221

221

00

sin gttV

tatvyy yy

−=

++=

θ

Express the projectile’s position vector :rr

( )[ ] ( )[ ] jir ˆsinˆcos 221 gttVtV −+= θθ

r

Using constant-acceleration equations, express the velocity of the projectile as a function of time:

θcos0 Vvv xx == and

gtVtavv yyy

−=

+=

θsin0

Express the projectile’s velocity vector :vr

[ ] [ ] jiv ˆsinˆcos gtVV −+= θθr

Substitute in equation (1) to obtain: ( )[ ] ( )[ ] [ ] [ ]

( )kji

jiL

ˆcos

ˆsinˆcos

ˆsinˆcos

221

221

θ

θθ

θθ

Vmgt

gtVVm

gttVtV

−=

−+×

−+=r

Differentiate L

rwith respect to t to obtain: ( )

( )k

kL

ˆcos

ˆcos221

θ

θ

mgtV

Vmgtdtd

dtd

−=

−=r

(2)

Using its definition, express the torque acting on the projectile:

( )( )[ ] ( )[ ]

( ) j

ji

jrτ

ˆ

ˆsinˆcos

ˆ2

21

mg

gttVtV

mg

−×

−+=

−×=

θθ

rr

or ( )kτ ˆcosθmgtV−=

r (3)

Comparing equations (2) and (3) we see that: τL rr

=dtd

Conservation of Angular Momentum *54 • Picture the Problem Let m represent the mass of the planet and apply the definition of torque to find the torque produced by the gravitational force of attraction. We can use Newton’s 2nd law of motion in the form dtdLτ

rr= to show that L

ris constant and apply

conservation of angular momentum to the motion of the planet at points A and B.

Chapter 10

760

(a) Express the torque produced by the gravitational force of attraction of the sun for the planet:

.ofdirection the

along acts because 0

r

FFrτr

rrrr=×=

(b) Because 0=τr : constant0 =×=⇒= vrLL rrr

r

mdtd

Noting that at points A and B

rv=× vr rr, express the

relationship between the distances from the sun and the speeds of the planets:

2211 vrvr =

or

1

2

2

1

rr

vv

=

55 •• Picture the Problem Let the system consist of you, the extended weights, and the platform. Because the net external torque acting on this system is zero, its angular momentum remains constant during the pulling in of the weights. (a) Using conservation of angular momentum, relate the initial and final angular speeds of the system to its initial and final moments of inertia:

ffii ωω II =

Solve for fω : i

f

if ωω

II

=

Substitute numerical values and evaluate fω : ( ) rev/s5.00rev/s1.5

mkg1.8mkg6

2

2

f =⋅

⋅=ω

(b) Express the change in the kinetic energy of the system:

2ii2

12ff2

1if ωω IIKKK −=−=∆

Substitute numerical values and evaluate ∆K: ( )

( )

J622

revrad2

srev1.5mkg6

revrad2

srev5mkg1.8

22

21

22

21

=

⎟⎠⎞

⎜⎝⎛ ×⋅−

⎟⎠⎞

⎜⎝⎛ ×⋅=∆

π

πK


761

(c) man. theofenergy internal thefrom

comesenergy thesystem, on the work doesagent external no Because

*56 •• Picture the Problem Let the system consist of the blob of putty and the turntable. Because the net external torque acting on this system is zero, its angular momentum remains constant when the blob of putty falls onto the turntable. (a) Using conservation of angular momentum, relate the initial and final angular speeds of the turntable to its initial and final moments of inertia and solve for ωf:

ffi0 ωω II =

and

if

0f ωω

II

=

Express the final rotational inertia of the turntable-plus-blob:

20blob0f mRIIII +=+=

Substitute and simplify to obtain: i

0

2i20

0f

1

1 ωωω

ImRmRI

I

+=

+=

(b) If the blob flies off tangentially to the turntable, its angular momentum doesn’t change (with respect to an axis through the center of turntable). Because there is no external torque acting on the blob-turntable system, the total angular momentum of the system will remain constant and the angular momentum of the turntable will not change. Because the moment of inertia of the table hasn’t changed either, the turntable will continue to spin at f ωω =' .

57 •• Picture the Problem Because the net external torque acting on the Lazy Susan-cockroach system is zero, the net angular momentum of the system is constant (equal to zero because the Lazy Susan is initially at rest) and we can use conservation of angular momentum to find the angular velocity ω of the Lazy Susan. The speed of the cockroach relative to the floor vf is the difference between its speed with respect to the Lazy Susan and the speed of the Lazy Susan at the location of the cockroach with respect to the floor. Relate the speed of the cockroach with respect to the floor vf to the speed of the Lazy Susan at the location of the cockroach:

rvv ω−=f (1)

Use conservation of angular momentum to obtain:

0CLS =− LL

Chapter 10

762

Express the angular momentum of the Lazy Susan:

ωω 221

LSLS MRIL ==

Express the angular momentum of the cockroach:

⎟⎠⎞

⎜⎝⎛ −== ωω

rvmrIL 2

CCC


21 =⎟

⎠⎞

⎜⎝⎛ −− ωω

rvmrMR

Solve for ω to obtain: 22 2

2mrMR

mrv+

=ω

Substitute in equation (1):

22

2

f 22

mrMRvmrvv

+−=

Substitute numerical values and evaluate vf:

( )( ) ( )( )( ) ( )( )

mm/s67.9m08.0kg015.02m15.0m25.0

m/s01.0m08.0kg0.0152m/s01.0 22

2

f =+

−=v

*58 •• Picture the Problem The net external torque acting on this system is zero and so we know that angular momentum is conserved as these disks are brought together. Let the numeral 1 refer to the disk to the left and the numeral 2 to the disk to the right. Let the angular momentum of the disk with the larger radius be positive. Using conservation of angular momentum, relate the initial angular speeds of the disks to their common final speed and to their moments of inertia:

ffii ωω II =

or ( ) f210201 ωωω IIII +=−

Solve for ωf: 0

21

21f ωω

IIII

+−

=

Express I1 and I2: ( ) 22

21

1 22 mrrmI ==

and 2

21

2 mrI =

Substitute and simplify to obtain:

053

02212

2212

f 22

ωωω =+−

=mrmrmrmr


763

59 •• Picture the Problem We can express the angular momentum and kinetic energy of the block directly from their definitions. The tension in the string provides the centripetal force required for the uniform circular motion and can be expressed using Newton’s 2nd law. Finally, we can use the work-kinetic energy theorem to express the work required to reduce the radius of the circle by a factor of two. (a) Express the initial angular momentum of the block:

000 mvrL =

(b) Express the initial kinetic energy of the block:

202

10 mvK =

(c) Using Newton’s 2nd law, relate the tension in the string to the centripetal force required for the circular motion:

0

20

c rvmFT ==

Use the work-kinetic energy theorem to relate the required work to the change in the kinetic energy of the block:

( ) 20

20

20

202

1

20

0f

20

0

20

f

20

0

20

f

2f

0f

321

2

1222

22

mrL

mrrmL

IIL

IL

IL

IL

ILKKKW

−=⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=−=

−=−=∆=

Substitute the result from part (a) and simplify to obtain:

203

2 mvW −=

*60 •• Picture the Problem Because the force exerted by the rubber band is parallel to the position vector of the point mass, the net external torque acting on it is zero and we can use the conservation of angular momentum to determine the speeds of the ball at points B and C. We’ll use mechanical energy conservation to find b by relating the kinetic and elastic potential energies at A and B. (a) Use conservation of momentum to relate the angular momenta at points A, B and C:

CBA LLL ==

or CCBBAA rmvrmvrmv ==

Solve for vB in terms of vA:

B

AAB r

rvv =

Chapter 10

764

Substitute numerical values and evaluate vB:

( ) m/s2.40m1m0.6m/s4 ==Bv

Solve for vC in terms of vA:

C

AAC r

rvv =

Substitute numerical values and evaluate vC:

( ) m/s00.4m6.0m0.6m/s4 ==Cv

(b) Use conservation of mechanical energy between points A and B to relate the kinetic energy of the point mass and the energy stored in the stretched rubber band:

BA EE =

or 2

212

212

212

21

BBAA brmvbrmv +=+

Solve for b: ( )22

22

BA

AB

rrvvmb

−−

=

Substitute numerical values and evaluate b: ( ) ( ) ( )[ ]

( ) ( )N/m20.3

m1m0.6m/s4m/s2.4kg2.0

22

22

=

−−

=b

Quantization of Angular Momentum *61 • Picture the Problem The electron’s spin angular momentum vector is related to its z component as shown in the diagram.

Using trigonometry, relate the magnitude of sr to its z component:

°== − 7.5475.0

cos 21

1

h

hθ

62 •• Picture the Problem Equation 10-27a describes the quantization of rotational energy. We can show that the energy difference between a given state and the next higher state is proportional to 1+l by using Equation 10-27a to express the energy difference.


765

From Equation 10-27a we have: ( ) r01 EK += lll

Using this equation, express the difference between one rotational state and the next higher state:

( )( ) ( )( ) r0

r0r0

12

121

E

EEE

+=

+−++=∆

l

llll

63 •• Picture the Problem The rotational energies of HBr molecule are related to l and

r0E according to ( ) r01 EK += lll where .22r0 IE h=

(a) Express and evaluate the moment of inertia of the H atom: ( ) ( )

247

2927

2p

mkg103.46

m100.144kg101.67

⋅×=

××=

=

−

−−

rmI

(b) Relate the rotational energies to l and r0E :

( ) r01 EK += lll

Evaluate r0E : ( )( )

meV0.996J101.60

eV1J101.59

mkg103.462sJ101.05

2

1922

247

2342

r0

=×

××=

⋅×⋅×

==

−−

−

−

IE h

Evaluate E for l = 1: ( )( ) meV1.99meV0.996111 =+=E

Evaluate E for l = 2: ( )( )

meV98.5

meV0.9961222

=

+=E

Evaluate E for l = 3: ( )( )

meV0.21

meV0.9961333

=

+=E

64 •• Picture the Problem We can use the definition of the moment of inertia of point particles to calculate the rotational inertia of the nitrogen molecule. The rotational energies of nitrogen molecule are related to l and r0E according

to ( ) r01 EK += lll where .22r0 IE h=

Chapter 10

766

(a) Using a rigid dumbbell model, express and evaluate the moment of inertia of the nitrogen molecule about its center of mass:

2N

2N

2N

i

2ii

2 rm

rmrmrmI

=

+== ∑

Substitute numerical values and evaluate I: ( )( )( )246

21127

mkg101.41

m105.5kg101.66142

⋅×=

××=−

−−I

(b) Relate the rotational energies to l and r0E :

( ) r01 EE += lll

Evaluate r0E : ( )( )

meV0.244J101.60

eV1J1091.3

mkg1041.12sJ101.05

2

1923

246

2342

r0

=×

××=

⋅×⋅×

==

−−

−

−

IE h

Substitute to obtain: ( ) meV1244.0 += lllE

*65 •• Picture the Problem We can obtain an expression for the speed of the nitrogen molecule by equating its translational and rotational kinetic energies and solving for v. Because this expression includes the moment of inertia I of the nitrogen molecule, we can use the definition of the moment of inertia to express I for a dumbbell model of the nitrogen molecule. The rotational energies of a nitrogen molecule depend on the quantum number l according to ( ) .2/12/ 22 IILE hlll +==

Equate the rotational kinetic energy of the nitrogen molecule in its l = 1 quantum state and its translational kinetic energy:

2N2

11 vmE = (1)

Express the rotational energy levels of the nitrogen molecule:

( )II

LE2

12

22 hlll

+==

For l = 1:

( )II

E22

1 2111 hh

=+

=


767


2N2

12

vmI

=h

Solve for v to obtain: Im

vN

22h= (2)

Using a rigid dumbbell model, express the moment of inertia of the nitrogen molecule about its center of mass:

2N

2N

2N

i

2ii 2 rmrmrmrmI =+== ∑

and 22

NN 2 rmIm =


rmrmv

N22

N

2

22 hh

==

Substitute numerical values and evaluate v:

( ) ( )m/s5.82

m105.5kg101.6641sJ10055.1

1127

34

=

××⋅×

= −−

−

v

Collision Problems 66 •• Picture the Problem Let the zero of gravitational potential energy be at the elevation of the rod. Because the net external torque acting on this system is zero, we know that angular momentum is conserved in the collision. We’ll use the definition of angular momentum to express the angular momentum just after the collision and conservation of mechanical energy to determine the speed of the ball just before it makes its perfectly inelastic collision with the rod. Use conservation of angular momentum to relate the angular momentum before the collision to the angular momentum just after the perfectly inelastic collision:

mvrLL

== if

Use conservation of mechanical energy to relate the kinetic energy of the ball just before impact to its initial potential energy:

0ifif =−+− UUKK

or, because Ki = Uf = 0, 0if =−UK

Letting h represent the distance the ghv 2=

Chapter 10

768

ball falls, substitute for if and UK and solve for v to

obtain: Substitute for v to obtain: ghmrL 2f =

Substitute numerical values and evaluate Lf:

( )( ) ( )( )sJ14.0

m1.2m/s9.812m0.9kg3.2 2f

⋅=

=L

*67 •• Picture the Problem Because there are no external forces or torques acting on the system defined in the problem statement, both linear and angular momentum are conserved in the collision and the velocity of the center of mass after the collision is the same as before the collision. Let the direction the blob of putty is moving initially be the positive x direction and toward the top of the page in the figure be the positive y direction. Using its definition, express the location of the center of mass relative to the center of the bar:

cm mMmdy

+= below the center of the bar.

Using its definition, express the velocity of the center of mass:

mMmvv+

=cm

Using the definition of L in terms of I and ω, express ω:

cm

cm

IL

=ω (1)

Express the angular momentum about the center of mass:

( )

mMmMvd

mMmddmv

ydmvL

+=⎟

⎠⎞

⎜⎝⎛

+−=

−= cmcm

Using the parallel axis theorem, express the moment of inertia of the system relative to its center of mass:

( )2cm

2cm

2121

cm ydmMyMLI −++=

Substitute for ycm and simplify to obtain:


769

( )( )

( ) ( )( )

( )

mMmMd

ML

mM

mMdmMML

mM

dmM

mM

dMmML

mMmdmMd

mmM

dMmML

mMmd

dmmM

mdMMLI

++=

+

++=

++

++=

+−+

++

+=

+−+

++=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

22

2

22

2

22

2

222

2

2

222

222

121

121

121

121

121

cm

Substitute for Icm and Lcm in equation (1) and simplify to obtain: ( ) 22

121 MmdmMML

mMvd++

=ω

Remarks: You can verify the expression for Icm by letting m → 0 to obtain

2cm MLI 12

1= and letting M → 0 to obtain Icm = 0. 68 •• Picture the Problem Because there are no external forces or torques acting on the system defined in the statement of Problem 67, both linear and angular momentum are conserved in the collision and the velocity of the center of mass after the collision is the same as before the collision. Kinetic energy is also conserved as the collision of the hard sphere with the bar is elastic. Let the direction the sphere is moving initially be the positive x direction and toward the top of the page in the figure be the positive y direction and v′ and V′ be the final velocities of the objects whose masses are m and M, respectively. Apply conservation of linear momentum to obtain:

fi pp =

or '' MVmvmv += (1)

Apply conservation of angular momentum to obtain:

fi LL =

or ω2

121' MLdmvmvd += (2)

Set v′ = 0 in equation (1) and solve for V ′: M

mvV' = (3)

Use conservation of mechanical energy to relate the kinetic energies of translation and rotation before

fi KK =

or ( ) 22121

212

212

21 ' ωMLMVmv += (4)

Chapter 10

770

and after the elastic collision: Substitute (2) and (3) in (4) and simplify to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+= 2

2121Ld

Mm

Mm

Solve for d:

mmMLd

12−

=

69 •• Picture the Problem Let the zero of gravitational potential energy be a distance x below the pivot as shown in the diagram. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing.

Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle θ:

0=∆+∆ UK or, because Kf = Ui = 0,

0fi =+− UK

and

( )θω cos12

221 −⎟

⎠⎞

⎜⎝⎛ += mgxdMgI (1)

Apply conservation of momentum to the collision:

fi LL =

or ( )[ ]ωω mdMdIdmv 22

31 8.08.0 +==

Solve for ω to obtain:

2231 64.0

8.0mdMd

dmv+

=ω (2)

Express the moment of inertia of the system about the pivot:

( )2

312

2312

64.0

8.0

Mdmd

MddmI

+=

+= (3)


771

Substitute equations (2) and (3) in equation (1) and simplify to obtain:

( )

( )22

31

2

64.032.0

cos12

mdMddmv

mgddMg

+=

−⎟⎠⎞

⎜⎝⎛ + θ

Solve for v:

( ) ( ) ( )2

2231

32.0cos164.08.05.0

dmgmdMdmMv θ−++

=

Evaluate v for θ = 90° to obtain:

( )( )2

2231

32.064.08.05.0

dmgmdMdmMv ++

=

70 •• Picture the Problem Let the zero of gravitational potential energy be a distance x below the pivot as shown in the diagram. Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. We can also use conservation of mechanical energy to relate the initial kinetic energy of the system after the collision to its potential energy at the top of its swing. Using conservation of mechanical energy, relate the rotational kinetic energy of the system just after the collision to its gravitational potential energy when it has swung through an angle θ :

0ifif =−+− UUKK

or, because Kf = Ui = 0, 0fi =+− UK

and

( )θω cos12

221 −⎟

⎠⎞

⎜⎝⎛ += mgxdMgI (1)

Apply conservation of momentum to the collision:

fi LL =

or

( )[ ]ωω

mdMd

Idmv22

31 8.0

8.0

+=

=

Chapter 10

772

Solve for ω to obtain: 22

21 64.0

8.0mdMd

dmv+

=ω (2)

Express the moment of inertia of the system about the pivot:

( )( )

( ) ( )[ ]( )2

231

231

2312

mkg0.660

m1.2kg0.8kg0.30.64

64.0

8.0

⋅=

+=

+=

+=

dMm

MddmI

Substitute equation (2) in equation (1) and simplify to obtain:

( )

( )Idmv

dmgdMg

232.0

cos18.02

=

−⎟⎠⎞

⎜⎝⎛ + θ

Solve for v: ( )( )

232.0cos18.05.0

dmImMgv θ−+

=

Substitute numerical values and evaluate v for θ = 60° to obtain:

( ) ( ) ( )[ ]( )( )( )( )

m/s74.7kg0.3m1.20.32

mkg0.6605.0kg0.30.8kg0.80.5m/s9.812

22

=⋅+

=v

71 •• Picture the Problem Let the length of the uniform stick be l. We can use the impulse-change in momentum theorem to express the velocity of the center of mass of the stick. By expressing the velocity V of the end of the stick in terms of the velocity of the center of mass and applying the angular impulse-change in angular momentum theorem we can find the angular velocity of the stick and, hence, the velocity of the end of the stick. (a) Apply the impulse-change in momentum theorem to obtain:

ppppK =−=∆= 0 or, because p0 = 0 and p = Mvcm,

cmMvK =

Solve for vcm to obtain: M

Kv =cm

(b) Relate the velocity V of the end of the stick to the velocity of the center of mass vcm:

( )l21

cmm of c torelcm ω+=+= vvvV (1)

Relate the angular impulse to the change in the angular momentum of the stick:

( ) ωcm021 ILLLK =−=∆=l

or, because L0 = 0, ( ) ωcm2

1 IK =l


773

Refer to Table 9-1 to find the moment of inertia of the stick with respect to its center of mass:

2121

cm lMI =

Substitute to obtain: ( ) ω2121

21 ll MK =

Solve for ω:

lMK6

=ω


MK

MK

MKV 4

26

=⎟⎠⎞

⎜⎝⎛+=

l

l

(c) Relate the velocity V′ of the other end of the stick to the velocity of the center of mass vcm:

( )

MK

MK

MK

vvvV

22

621

cmm of c torelcm

−=⎟⎠⎞

⎜⎝⎛−=

−=−=

l

l

lω

(d) Letting x be the distance from the center of mass toward the end not struck, express the condition that the point at x is at rest:

0cm =− xv ω

Solve for x to obtain: 06

=− xM

KMK

l

Solve for x to obtain:

l

l

61

6 ==

MK

MK

x

Note that for a meter stick struck at the

100-cm mark, the stationary point would be at the 33.3-cm mark.

Remarks: You can easily check this result by placing a meterstick on the floor and giving it a sharp blow at the 100-cm mark. 72 •• Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. (a) Use its definition to express the total angular momentum of the disk and projectile just before impact:

bvmL 0p0 =

Chapter 10

774

(b) Use conservation of angular momentum to relate the angular momenta just before and just after the collision:

ωILL ==0 and I

L0=ω

Express the moment of inertia of the disk + projectile:

2p

221 bmMRI +=

Substitute for I in the expression for ω to obtain: 2

p2

0p

22

bmMRbvm

+=ω

(c) Express the kinetic energy of the system after impact in terms of its angular momentum:

( )( )

( )2

p2

20p

2p

221

20p

2

f

2

22

bmMRbvm

bmMRbvm

ILK

+=

+==

(d) Express the difference between the initial and final kinetic energies, substitute, and simplify to obtain:

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−=

+−=

−=∆

2p

2

2p2

0p21

2p

2

20p2

0p21

fi

21

2

bmMRbm

vm

bmMRbvm

vm

KKE

*73 •• Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational kinetic of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the masses M and m. Use conservation of energy to relate the initial and final potential energy of the rod to its rotational kinetic energy just before it collides with the particle:

0ifif =−+− UUKK

or, because Ki = 0, 0iff =−+ UUK


775

Substitute for Kf, Uf, and Ui to obtain:

( ) 02 1

12213

121 =−+ MgLLMgML ω

Solve for ω:

1

3Lg

=ω

Letting ω′ represent the angular speed of the rod-and-particle system just after impact, use conservation of angular momentum to relate the angular momenta before and after the collision:

fi LL =

or ( ) ( ) '2

2213

1213

1 ωω mLMLML +=

Solve for ω′: ωω 2

2213

1

213

1

'mLML

ML+

=

Use conservation of energy to relate the rotational kinetic energy of the rod-plus-particle just after their collision to their potential energy when they have swung through an angle θmax:

0ifif =−+− UUKK

or, because Kf = 0, ( )( )

( ) 0cos1cos1'

max2

max1212

21

=−+

−+−

θθω

mgLLMgI

(1)

Express the moment of inertia of the system with respect to the pivot:

22

213

1 mLMLI +=

Substitute for θmax, I and ω′ in equation (1):

( )( ) 212

122

213

1

2213

1

1

3mgLLMg

mLML

MLLg

+=+

Simplify to obtain: 3

21222

21

31 632 L

MmLLLL

MmL ++= (2)

Simplify equation (2) by letting α = m/M and β = L2/L1 to obtain:

01236 232 =−++ αβββα

Substitute for α and simplify to obtain the cubic equation in β:

034912 23 =−++ βββ

Use the solver function* of your calculator to find the only real value

349.0=β

Chapter 10

776

of β:

*Remarks: Most graphing calculators have a ″solver″ feature. One can solve the cubic equation using either the ″graph″ and ″trace″ capabilities or the ″solver″ feature. The root given above was found using SOLVER on a TI-85. 74 •• Picture the Problem Because the net external torque acting on the system is zero, angular momentum is conserved in this perfectly inelastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the perfectly inelastic collision with the particle and the rotational kinetic energy of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the mass M to m. (a) Use conservation of energy to relate the initial and final potential energy of the rod to its rotational kinetic energy just before it collides with the particle:

0ifif =−+− UUKK


Substitute for Kf, Uf, and Ui to obtain:

( ) 02 1

12213


Solve for ω:

1

3Lg

=ω

Letting ω′ represent the angular speed of the system after impact, use conservation of angular momentum to relate the angular momenta before and after the collision:

fi LL =

or ( ) ( ) '2

2213

1213

1 ωω mLMLML += (1)

Solve for ω′:

122

213

1

213

1

22

213

1

213

1

3

'

Lg

mLMLML

mLMLML

+=

+= ωω


777

Substitute numerical values to obtain: ( )( )( )( ) ( )

( )

( )

m

m

m

64.0kg0.960s/kg75.4

m64.0mkg0.960s/mkg75.4

m2.1m/s81.93

m8.0m2.1kg2m2.1kg2'

22

2

2

2231

231

+=

+⋅⋅

=

×

+=ω

Use conservation of energy to relate the rotational kinetic energy of the rod-plus-particle just after their collision to their potential energy when they have swung through an angle θmax:

0ifif =−+− UUKK

or, because Kf = 0, 0ifi =−+− UUK

Substitute for Ki, Uf, and Ui to obtain:

( )( )( ) 0cos1

cos1'

max2

max1212

21

=−+−+−

θθω

mgLLMgI

Express the moment of inertia of the system with respect to the pivot:

22

213

1 mLMLI +=

Substitute for θmax, I and ω′ in equation (1) and simplify to obtain:

( ) ( )21

221

2.064.0kg960.0

kg/s75.4 mLMLgm

+=+

Substitute for M, L1 and L2 and simplify to obtain:

0901.800.32 =−+ mm

Solve the quadratic equation for its positive root:

kg84.1=m

(b) The energy dissipated in the inelastic collision is:

fi UUE −=∆ (2)

Express Ui: 2

1i

LMgU =

Express Uf: ( ) ⎟

⎠⎞

⎜⎝⎛ +−= 2

1maxf 2

cos1 mLLMgU θ

Chapter 10

778


( ) ⎟⎠⎞

⎜⎝⎛ +−−

=∆

21

max

1

2cos1

2

mLLMg

LMgE

θ

Substitute numerical values and evaluate ∆E:

( )( )( )

( )( ) ( )( ) ( )( )

J51.6

m0.8kg1.852

m1.2kg2m/s9.81cos371

2m1.2m/s9.81kg2

2

2

f

=

⎟⎠⎞

⎜⎝⎛ +°−−

=U

75 •• Picture the Problem Let ωi and ωf be the angular velocities of the rod immediately before and immediately after the inelastic collision with the mass m. Let ω0 be the initial angular velocity of the rod. Choose the zero of gravitational potential energy be at a distance L1 below the pivot. We apply energy conservation to determine ωf and conservation of angular momentum to determine ωi. We’ll apply energy conservation to determine ω0. Finally, we’ll find the energies of the system immediately before and after the collision and the energy dissipated. Express the energy dissipated in the inelastic collision:

fi UUE −=∆ (1)

Use energy conservation to relate the kinetic energy of the system immediately after the collision to its potential energy after a 180° rotation:

0ifif =−+− UUKK

or, because Kf = Ktop = 0 and Ki = Kbottom, 0bottomtopbottom =−+− UUK

Substitute for Kbottom, Utop, and Ubottom to obtain:

( )( ) 02112

1

211232

f21

=−−−+++−LLmgMgLLLmgMgLIω

Simplify to obtain:

02 212f2

1 =++− mgLMgLIω (2)

Express I: 22

213

1 mLMLI +=

Substitute for I in equation (2) and solve for ωf to obtain:

( )22

213

121

f22mLML

mLMLg++

=ω


779

Substitute numerical values and evaluate ωf:

( ) ( )( ) ( )( )[ ]( )( ) ( )( )

rad/s00.7m0.8kg0.4m1.2kg0.75

m0.8kg0.42m1.2kg0.75m/s9.81222

31

2

f =+

+=ω

Use conservation of angular momentum to relate the angular momentum of the system just before the collision to its angular momentum just after the collision:

fi LL =

or ffii ωω II =

Substitute for Ii and If and solve for ωi:

( ) ( ) f22

213

1i

213

1 ωω mLMLML +=

and

f

2

1

2i

31 ωω⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

LL

Mm

Substitute numerical values and evaluate ωi:

( ) ( )

rad/s0.12

rad/s7.00m1.2m0.8

kg0.75kg0.431

2

i

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=ω

Apply conservation of mechanical energy to relate the initial rotational kinetic energy of the rod to its rotational kinetic energy just before its collision with the particle:

0ifif =−+− UUKK

Substitute to obtain: ( ) ( )0

21

120

213

1212

i213

121

=−

+−

MgL

LMgMLML ωω

Solve for ω0:

1

2i0

3Lg

−= ωω

Substitute numerical values and evaluate ω0: ( ) ( )

rad/s10.9

m1.2m/s9.813

rad/s122

20

=

−=ω

Chapter 10

780

Substitute in equation (1) to express the energy dissipated in the collision:

( ) 212i

213

121 2mgLMgLMLE +−=∆ ω


( )( ) ( ) ( ) ( )( ) ( )( )[ ]J8.10

m0.8kg0.42m1.2kg0.75m/s9.81rad/s12m1.2kg0.75 22261

=

+−=∆E

76 ••• Picture the Problem Let v be the speed of the particle immediately after the collision and ωi and ωf be the angular velocities of the rod immediately before and immediately after the elastic collision with the mass m. Choose the zero of gravitational potential energy be at a distance L1 below the pivot. Because the net external torque acting on the system is zero, angular momentum is conserved in this elastic collision. The rod, on its downward swing, acquires rotational kinetic energy. Angular momentum is conserved in the elastic collision with the particle and the kinetic energy of the after-collision system is then transformed into gravitational potential energy as the rod-plus-particle swing upward. Let the zero of gravitational potential energy be at a distance L1 below the pivot and use both angular momentum and mechanical energy conservation to relate the distances L1 and L2 and the mass M to m. Use energy conservation to relate the energies of the system immediately before and after the elastic collision:

0ifif =−+− UUKK


Substitute for Kf, Uf, and Ui to obtain: ( ) 02

cos12

1max

1221 =−−+

LMg

LMgmv θ

Solve for mv2: max1

2 cosθMgLmv = (1)

Apply conservation of energy to express the angular speed of the rod just before the collision:

0ifif =−+− UUKK


Substitute for Kf, Uf, and Ui to obtain: ( ) 0

2 112

i213


Solve for ωi:

1i

3Lg

=ω


781

Apply conservation of energy to the rod after the collision:

( ) ( ) 0cos12 max

12f

213

121 =−− θω LMgML

Solve for ωf:

1f

6.0L

g=ω

Apply conservation of angular momentum to the collision:

fi LL =

or ( ) ( ) 2f

213

1i

213

1 mvLMLML += ωω

Solve for mv: ( )

2

fi213

1

LMLmv ωω −

=

Substitute for ωf and ωI to obtain:

2

11

21

3

6.03

LL

gLgML

mv⎟⎟⎠

⎞⎜⎜⎝

⎛−

= (2)

Divide equation (1) by equation (2) to eliminate m and solve for v:

11

max2

2

11

21

max1

6.03cos3

3

6.03cos

gLgLgL

LL

gLgML

MgLv

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

θ

θ


( ) ( )( ) ( ) ( )( )

m/s72.5m2.1m/s81.96.0m2.1m/s81.93

37cosm8.0m/s81.9322

2

=−

°=v

Solve equation (1) for m:

2max1 cos

vMgLm θ

=

Substitute for v in the expression for mv and solve for m:

( )( )( )( )

kg575.0

m/s72.537cosm2.1m/s81.9kg2

2

2

=

°=m

Because the collision was elastic: 0=∆E

Chapter 10

782

77 •• Picture the Problem We can determine the angular momentum of the wheel and the angular velocity of its precession from their definitions. The period of the precessional motion can be found from its angular velocity and the angular momentum associated with the motion of the center of mass from its definition. (a) Using the definition of angular momentum, express the angular momentum of the spinning wheel:

ωωω 22 RgwMRIL ===


( )

sJ1.18

revrad2

srev12

m0.28m/s9.81N30 2

2

⋅=

⎟⎠⎞

⎜⎝⎛ ××

⎟⎟⎠

⎞⎜⎜⎝

⎛=

π

L

(b) Using its definition, express the angular velocity of precession: L

MgDdtd

==φωp

Substitute numerical values and evaluate ωp:

( )( ) rad/s0.414sJ18.1

m0.25N30p =

⋅=ω

(c) Express the period of the precessional motion as a function of the angular velocity of precession:

s2.15rad/s414.0

22

p

===π

ωπT

(d) Express the angular momentum of the center of mass due to the precession:

p2

pcmp ωω MDIL ==

Substitute numerical values and evaluate Lp:

( ) ( )

sJ0791.0

rad/s414.0m0.25m/s9.81N30 2

2p

⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛=L

The direction of Lp is either up or down, depending on the direction of L.

*78 •• Picture the Problem The angular velocity of precession can be found from its definition. Both the speed and acceleration of the center of mass during precession are related to the angular velocity of precession. We can use Newton’s 2nd law to find the vertical and


783

horizontal components of the force exerted by the pivot. (a) Using its definition, express the angular velocity of precession:

s2

s2

21

ssp

2ωωω

φωR

gDMRMgD

IMgD

dtd

====

Substitute numerical values and evaluate ωp:

( ) ( )

( )rad/s27.3

s60min1

revrad2π

minrev700m0.064

m0.05m/s9.8122

2

p =

⎟⎟⎠

⎞⎜⎜⎝

⎛××

=ω

(b) Express the speed of the center of mass in terms of its angular velocity of precession:

( )( )m/s0.164

rad/s3.27m0.05pcm

=

== ωDv

(c) Relate the acceleration of the center of mass to its angular velocity of precession:

( )( )2

22pcm

m/s0.535

rad/s3.27m0.05

=

== ωDa

(d) Use Newton’s 2nd law to relate the vertical component of the force exerted by the pivot to the weight of the disk:

( )( )N24.5

m/s9.81kg2.5 2v

=

== MgF

Relate the horizontal component of the force exerted by the pivot to the acceleration of the center of mass:

( )( )N34.1

m/s535.0kg2.5 2cmv

=

== MaF

General Problems 79 • Picture the Problem While the 3-kg particle is moving in a straight line, it has angular momentum given by prL

rrr×= where r

ris its position vector and p

ris its linear

momentum. The torque due to the applied force is given by .Frτrrr

×= (a) Express the angular momentum of the particle:

prLrrr

×=

Express the vectors rr and pr : ( ) ( ) jir ˆm3.5ˆm12 +=r

Chapter 10

784

and ( )( )

( )i

iipˆm/skg9

ˆm/s3kg3ˆ

⋅=

== mvr

Substitute and simplify to find L

r: ( ) ( )[ ] ( )

( )( )( )k

ij

ijiL

ˆ/smkg7.47

ˆˆ/smkg7.47

ˆm/skg9ˆm3.5ˆm12

2

2

⋅−=

×⋅=

⋅×+=r

(b) Using its definition, express the torque due to the force:

Frτrrr

×=

Substitute and simplify to find τr : ( ) ( )[ ] ( )( )( )( )k

ij

ijiτ

ˆmN9.15

ˆˆmN9.15

ˆN3ˆm3.5ˆm12

⋅=

×⋅−=

−×+=r

80 • Picture the Problem The angular momentum of the particle is given by

prLrrr

×= where rr

is its position vector and pr

is its linear momentum. The torque acting

on the particle is given by .dtdLτrr

=

Express the angular momentum of the particle:

dtdm

mmrr

vrvrprLr

r

rrrrrrr

×=

×=×=×=

Evaluate dtdrr

: jr ˆ6tdtd

=r

Substitute and simplify to find L

r: ( ) ( ) ( ) [ ]

( )( )k

j

jiL

ˆsJ0.72

ˆm/s6

ˆm/s3ˆm4kg3 22

⋅=

×

+=

t

t

tr

Find the torque due to the force: ( )[ ]

( )k

kLτ

ˆmN0.72

ˆsJ0.72

⋅=

⋅== tdtd

dtdr

r


785

81 •• Picture the Problem The ice skaters rotate about their center of mass; a point we can locate using its definition. Knowing the location of the center of mass we can determine their moment of inertia with respect to an axis through this point. The angular momentum of the system is then given by ωcmIL = and its kinetic energy can be found

from .2 cm2 ILK =

(a) Express the angular momentum of the system about the center of mass of the skaters:

ωcmIL =

Using its definition, locate the center of mass, relative to the 85-kg skater, of the system:

( )( ) ( )( )

m0.668kg85kg55

0kg85m1.7kg55cm

=+

+=x

Calculate cmI : ( )( )( )( )

2

2

2cm

mkg5.96m0.668kg85

m0.668m1.7kg55

⋅=

+

−=I

Substitute to determine L: ( )

sJ243

revrad2π

s2.5rev1mkg96.5 2

⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛×⋅=L

(b) Relate the total kinetic energy of the system to its angular momentum and evaluate K:

cm

2

2ILK =

Substitute numerical values and evaluate K:

( )( ) J306

mkg96.52sJ243

2

2

=⋅

⋅=K

Chapter 10

786

*82 •• Picture the Problem Let the origin of the coordinate system be at the pivot (point P). The diagram shows the forces acting on the ball. We’ll apply Newton’s 2nd law to the ball to determine its speed. We’ll then use the derivative of its position vector to express its velocity and the definition of angular momentum to show that L

rhas

both horizontal and vertical components. We can use the derivative of L

rwith

respect to time to show that the rate at which the angular momentum of the ball changes is equal to the torque, relative to the pivot point, acting on it. (a) Express the angular momentum of the ball about the point of support:

vrprLrrrrr

×=×= m (1)

Apply Newton’s 2nd law to the ball: ∑ ==θ

θsin

sin2

rvmTFx

and

∑ =−= 0cos mgTFz θ

Eliminate T between these equations and solve for v:

θθ tansinrgv =


( )( )m/s2.06

tan30sin30m/s9.81m1.5 2

=

°°=v

Express the position vector of the ball:

( ) ( )( ) k

jirˆ30cosm5.1

ˆsinˆcos30sinm5.1

°−

+°= tt ωωr

where .kωω =

Find the velocity of the ball:

( )( )ji

rv

ˆcosˆsinm/s75.0 ttdtd

ωωω +−=

=r

r

Evaluate ω:

( ) rad/s75.230sinm5.1

m/s06.2=

°=ω


787

Substitute for ω to obtain:

( )( )jiv ˆcosˆsinm/s06.2 tt ωω +−=r

Substitute in equation (1) and evaluate Lr

:

( ) ( ) ( ) ( )[ ]( )[ ( )]

( )[ ] sJˆ09.3ˆsinˆcos36.5

ˆcosˆsinm/s06.2

ˆ30cosm5.1ˆsinˆcos30sinm5.1kg2

⋅++=

+−×

°−+°=

kji

ji

kjiL

tt

tt

tt

ωω

ωω

ωωr

The horizontal component of L

ris:

( ) sJˆsinˆcos36.5 ⋅+ ji tt ωω

The vertical component of Lr

is:

sJˆ09.3 ⋅k

(b) Evaluate dtdLr

: ( )[ ] Jˆcosˆsin36.5 jiL ttdtd ωωω +−=r

Evaluate the magnitude of dtdLr

: ( )( )

mN7.14

rad/s75.2smN36.5

⋅=

⋅⋅=dtdLr

Express the magnitude of the torque exerted by gravity about the point of support:

θτ sinmgr=

Substitute numerical values and evaluate τ :

( )( )( )mN7.14

30sinm1.5m/s9.81kg2 2

⋅=

°=τ

83 •• Picture the Problem In part (a) we need to decide whether a net torque acts on the object. In part (b) the issue is whether any external forces act on the object. In part (c) we can apply the definition of kinetic energy to find the speed of the object when the unwrapped length has shortened to r/2. (a) Consider the overhead view of the cylindrical post and the object shown in the adjoining figure. The object rotates counterclockwise. The torque about the center of the cylinder is clockwise and of magnitude RT, where R is the radius of the cylinder and T is the tension. So

Chapter 10

788

L must decrease.

decreases. No, L

(b) Because, in this frictionless environment, no net external forces act on the object:

constant. isenergy kinetic Its

(c) Express the kinetic energy of the object as it spirals inward:

( ) 221

2

22

212

21 mv

rvmrIK === ω

constant.)remainsenergy kinetic (The .0v

84 •• Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular velocities of the system. Using conservation of angular momentum, relate the initial and final angular velocities to the initial and final moments of inertia:

fi LL =

or ffii ωω II =

Solve for ωi: ωωωf

ii

f

if I

III

==

Express Ii: ( )2

412

101

i 2 lmMLI +=

Express If: ( )2

412

101

f 2 mLMLI +=

Substitute to express fω in terms of ω : ( )

( )

ω

ωω

mML

mM

mLMLmML

5

5

22

2

2

2412

101

2412

101

f

+

+=

++

=

l

l

Express the initial kinetic energy of the system:

( )[ ]( ) 222

201

22412

101

212

i21

i

5

2

ω

ωω

l

l

mML

mMLIK

+=

+==


789

Express the final kinetic energy of the system and simplify to obtain:

( )[ ] ( )

( )

( ) 222

222

201

2

22

201

2

2

2

22201

2f

222012

f2

412

101

212

ff21

f

55

5

5

5

55

52

ω

ωω

ωωω

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++

=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+

++=

+=+==

mLMLmML

mML

mML

mML

mMmLML

mLMLmLMLIK

l

ll

85 •• Determine the Concept Yes. The net external torque is zero and angular momentum is conserved as the system evolves from its initial to its final state. Because the disks come to the same final position, the initial and final configurations are the same as in Problem 84. Therefore, the answers are the same as for Problem 84. 86 •• Picture the Problem Because the net torque acting on the system is zero; we can use conservation of angular momentum to relate the initial and final angular velocities of the system. Using conservation of angular momentum, relate the initial and final angular velocities to the initial and final moments of inertia:

fi LL =

or ffii ωω II =

Solve for ωf: ωωωf

ii

f

if I

III

== (1)

Relate the tension in the string to the angular speed of the system and solve for and evaluate ω:

22

2ωω lmmrT ==

and ( )

( )( )rad/s30.0

m0.6kg0.4N10822

=

==lm

Tω

Chapter 10

790

Express and evaluate Ii: ( )( )( ) ( )( )

2

2212

101

2412

101

i

mkg0.392

m0.6kg0.4m2kg0.8

2

⋅=

+=

+= lmMLI

Express and evaluate If: ( )

( )( ) ( )( )2

2212

101

2412

101

f

mkg12.1

m2kg0.4m2kg0.8

2

⋅=

+=

+= mLMLI

Substitute in equation (1) and solve for fω : ( )

rad/s5.10

rad/s0.30mkg1.12mkg392.0

2

2

f

if

=

⋅⋅

== ωωII

Express and evaluate the initial kinetic energy of the system: ( )( )

J176

rad/s0.30mkg392.0 2221

2i2

1i

=

⋅=

= ωIK

Express and evaluate the final kinetic energy of the system: ( )( )

J7.61

rad/s10.5mkg1.12 2221

2ff2

1f

=

⋅=

= ωIK

87 •• Picture the Problem Until the inelastic collision of the cylindrical objects at the ends of the cylinder, both angular momentum and energy are conserved. Let K’ represent the kinetic energy of the system just before the disks reach the end of the cylinder and use conservation of energy to relate the initial and final kinetic energies to the final radial velocity. Using conservation of mechanical energy, relate the initial and final kinetic energies of the disks:

'i KK =

or ( )2

r212

ff212

i21 2mvII += ωω

Solve for vr:

mII

v2

2ff

2i

rωω −

= (1)

Using conservation of angular momentum, relate the initial and final angular velocities to the initial

fi LL =

or


791

and final moments of inertia:

ffii ωω II =

Solve for fω : ωωωf

i

f

if I

III

==

Express Ii: ( )2

412

101

i 2 lmMLI +=

Express If: ( )2

412

101

f 2 mLMLI +=

Substitute to obtain fω in terms of ω : ( )

( )ω

ωω

22

22

2412

101

2412

101

f

55

22

mLMLmML

mLMLmML

++

=

++

=

l

l

Substitute in equation (1) and simplify to obtain:

( )22r 2

ll

−= LL

v ω

88 •• Picture the Problem Because the net torque acting on the system is zero, we can use conservation of angular momentum to relate the initial and final angular velocities and the initial and final kinetic energy of the system. Using conservation of angular momentum, relate the initial and final angular velocities to the initial and final moments of inertia:

fi LL =

or ffii ωω II =

Solve for fω : ωωωf

ii

f

if I

III

== (1)

Relate the tension in the string to the angular speed of the system:

22

2ωω lmmrT ==

Solve for ω:

lmT2

=ω


( )( )( ) rad/s30.0

m0.6kg0.4N1082

==ω

Chapter 10

792

Express and evaluate Ii: ( )( )( ) ( )( )

2

2212

101

2412

101

i

mkg0.392

m0.6kg0.4m2kg0.8

2

⋅=

+=

+= lmMLI

Letting L′ represent the final separation of the disks, express and evaluate If:

( )( )( ) ( )( )

2

2212

101

2412

101

f

mkg832.0

m6.1kg0.4m2kg0.8

'2

⋅=

+=

+= mLMLI

Substitute in equation (1) and solve for fω : ( )

rad/s1.14

rad/s0.30mkg832.0mkg392.0

2

2

f

if

=⋅⋅

== ωωII

Express and evaluate the initial kinetic energy of the system: ( )( )

J176

rad/s0.30mkg392.0 2221

2i2

1i

=

⋅=

= ωIK

Express and evaluate the final kinetic energy of the system: ( )( )

J7.82

rad/s1.41mkg832.0 2221

2ff2

1f

=

⋅=

= ωIK

The energy dissipated in friction is:

J93.3

J82.7J176fi

=

−=−=∆ KKE

*89 •• Picture the Problem The drawing shows an elliptical orbit. The triangular element of the area is ( ) .2

21

21 θθ drrdrdA ==

Differentiate dA with respect to t to obtain:

ωθ 2212

21 r

dtdr

dtdA

==

Because the gravitational force acts along the line joining the two objects, τ = 0 and:

constant

2

== ωmrL


793

Eliminate r2ω between the two equations to obtain:

constant2

==mL

dtdA

90 •• Picture the Problem Let x be the radial distance each disk moves outward. Because the net torque acting on the system is zero, we can use conservation of angular momentum to relate the initial and final angular velocities to the initial and final moments of inertia. We’ll assume that the disks are thin enough so that we can ignore their lengths in expressing their moments of inertia. Use conservation of angular momentum to relate the initial and final angular velocities of the disks:

fi LL =

or ffii ωω II =

Solve for ωf: i

f

if ωω

II

= (1)

Express the initial moment of inertia of the system:

diskcyli 2III +=

Express the moment of inertia of the cylinder: ( )

( ) ( ) ( )[ ]2

22121

22121

2212

121

cyl

mkg0.232

m0.26m1.8kg0.8

6

⋅=

+=

+=

+=

RLM

MRMLI

Letting l represent the distance of the clamped disks from the center of rotation and ignoring the thickness of each disk (we’re told they are thin), use the parallel-axis theorem to express the moment of inertia of each disk:

( )( ) ( ) ( )[ ]

2

2241

2241

2241

disk

mkg0340.0

m4.04m2.0kg2.0

4

⋅=

+=

+=

+=

l

l

rm

mmrI

With the disks clamped:

( )2

22

diskcyli

mkg300.0mkg0340.02mkg232.0

2

⋅=

⋅+⋅=

+= III

Chapter 10

794

With the disks unclamped, l = 0.6 m and:

( )( ) ( ) ( )[ ]

2

2241

2241

disk

mkg0740.0

m6.04m2.0kg2.0

4

⋅=

+=

+= lrmI

Express and evaluate the final moment of inertia of the system: ( )

2

22

diskcylf

mkg380.0mkg0740.02mkg232.0

2

⋅=

⋅+⋅=

+= III

Substitute in equation (1) to determine ωf:

( )

rad/s32.6

rad/s8mkg380.0mkg300.0

2

2

f

=

⋅⋅

=ω

Express the energy dissipated in friction: ( )2

212

ff212

ii21

fi

kxII

EEE

+−=

−=∆

ωω

Apply Newton’s 2nd law to each disk when they are in their final positions:

∑ == 2radial ωmrkxF

Solve for k: x

mrk2ω

=

Substitute numerical values and evaluate k:

( )( )( )

N/m24.0m0.2

rad/s6.32m0.6kg0.2 2

=

=k

Express the energy dissipated in friction:

( )2212

ff212

ii21

fifr

kxII

EEW

+−=

−=

ωω

Substitute numerical values and evaluate Wfr:

( )( ) ( )( ) ( )( )J1.53

m0.2N/m24rad/s32.6mkg380.0rad/s8mkg0.300 22122

2122

21

fr

=

−⋅−⋅=W

91 •• Picture the Problem Let the letters d, m, and r denote the disk and the letters t, M, and R the turntable. We can use conservation of angular momentum to relate the final angular speed of the turntable to the initial angular speed of the Euler disk and the moments of inertia of the turntable and the disk. In part (b) we’ll need to use the parallel-axis theorem


795

to express the moment of inertia of the disk with respect to the rotational axis of the turntable. You can find the moments of inertia of the disk in its two orientations and that of the turntable in Table 9-1. (a) Use conservation of angular momentum to relate the initial and final angular momenta of the system:

tftfdfdfdidi ωωω III +=

Because ωtf = ωdf:

tftftfdfdidi ωωω III +=

Solve for ωtf: di

tfdf

ditf ωω

III+

= (1)

Ignoring the negligible thickness of the disk, express its initial moment of inertia:

241

di mrI =

Express the final moment of inertia of the disk:

221

df mrI =

Express the final moment of inertia of the turntable:

221

tf MRI =


di

2

2

di2212

21

241

tf

22

1 ω

ωω

mrMR

MRmrmr

+=

+=

(2)

Express ωdi in rad/s:

rad/ss60

min1rev

rad2minrev30di ππω =××=

Substitute numerical values in equation (2) and evaluate ωtf: ( )( )

( )( )rad/s228.0

m0.125kg0.5m0.25kg0.73522

rad/s

2

2tf

=

+=

πω

(b) Use the parallel-axis theorem to express the final moment of inertia of the disk when it is a distance L from the center of the turntable:

( )222122

21

df LrmmLmrI +=+=

Chapter 10

796


( )di

2

2

2

2

di22122

21

241

tf

242

1 ω

ωω

mrMR

rL

MRLrmmr

++=

++=

Substitute numerical values and evaluate ωtf:

( )( )

( )( )( )( )

rad/s192.0

m0.125kg0.5m0.25kg0.7352

m0.125m0.142

rad/s

2

2

2

2tf =++

=πω

92 •• Picture the Problem We can express the period of the earth’s rotation in terms of its angular velocity of rotation and relate its angular velocity to its angular momentum and moment of inertia with respect to an axis through its center. We can differentiate this expression with respect to T and then use differentials to approximate the changes in r and T. (a) Express the period of the earth’s rotation in terms of its angular velocity of rotation:

ωπ2

=T

Relate the earth’s angular velocity of rotation to its angular momentum and moment of inertia:

252 mr

LIL

==ω

Substitute and simplify to obtain: ( ) 22

52

542

rLm

Lmr

T ππ==

(b) Find dT/dr:

rTr

rTr

Lm

drdT 22

542 2 =⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

π

Solve for dT/T:

rdr

TdT 2= or

rr

TT ∆

≈∆ 2

(c) Using the equation we just derived, substitute for the change in the period of the earth:

rr

TT ∆

==×=∆ 2

14601

d365.24y1

yd4

1


797

Solve for and evaluate ∆r: ( ) ( )

km18.2

14602km1037.6

14602

3

=

×==∆

rr

*93 •• Picture the Problem Let ωP be the angular velocity of precession of the earth-as-gyroscope, ωs its angular velocity about its spin axis, and I its moment of inertia with respect to an axis through its poles, and relate ωP to ωs and I using its definition. Use its definition to express the precession rate of the earth as a giant gyroscope:

Lτω =P

Substitute for I and solve for τ: PP ωωωτ IL ==

Express the angular velocity ωs of the earth about its spin axis: T

πω 2= where T is the period of rotation of

the earth.

Substitute to obtain: T

I P2 ωπτ =

Substitute numerical values and evaluateτ:

( ) ( ) mN1047.4

hs3600

dh24d1

s1066.7mkg1003.82 22112237

⋅×=××

×⋅×=

−−πτ

94 ••• Picture the Problem The applied torque accelerates the system and increases the tension in the string until it breaks. The work done before the string breaks is the change in the kinetic energy of the system. We can use Newton’s 2nd law to relate the breaking tension to the angular velocity of the system at the instant the string breaks. Once the applied torque is removed, angular momentum is conserved. Express the work done before the string breaks:

2ff2

1f ωIKKW ==∆= (1)

Express the moment of inertia of the system (see Table 9-1):

( )( )( ) ( )

( ) 22

22121

22cylcyl12

1mcyl

kg0.8mkg256.0

kg4.02m1.6kg1.2

22

x

x

mxLMxIIII

+⋅=

+=

+==+=

Chapter 10

798

Evaluate If = I(0.4 m): ( )( )( )

2

22

f

mkg384.0m4.0kg0.8mkg256.0

m4.0

⋅=

+⋅=

= II

Using Newton’s 2nd law, relate the forces acting on a disk to its angular velocity:

∑ == 2fradial ωmrTF

where T is the tension in the string at which it breaks.

Solve for ωf:

mrT

=fω

Substitute numerical values and evaluate ωf: ( )( ) rad/s0.25

m0.4kg0.4N100

f ==ω

Substitute in equation (1) to express the work done before the string breaks:

2ff2

1 ωIW =

Substitute numerical values and evaluate W:

( )( )J120

rad/s25mkg0.384 2221

=

⋅=W

With the applied torque removed, angular momentum is conserved and we can express the angular momentum as a function of x:

( ) ( )xxIIL

ωω

== ff

Solve for ( )xω : ( ) ( )xIIx ffωω =

Substitute numerical values to obtain: ( ) ( )( )

( )

( ) 22

22

2

kg0.8mkg0.256sJ60.9

kg0.8mkg0.256rad/s25mkg0.384

x

xx

+⋅⋅

=

+⋅⋅

=ω

95 ••• Picture the Problem The applied torque accelerates the system and increases the tension in the string until it breaks. The work done before the string breaks is the change in the kinetic energy of the system. We can use Newton’s 2nd law to relate the breaking tension to the angular velocity of the system at the instant the string breaks. Once the applied


799

torque is removed, angular momentum is conserved. Express the work done before the string breaks:

2ff2

1f ωIKKW ==∆= (1)

Express the moment of inertia of the system (see Table 9-1):

( ) 22cylcyl12

1mcyl 22 mxLMxIIII +==+=

Substitute numerical values to obtain:

( )( ) ( )( ) 22

22121

kg0.8mkg256.0

kg4.02m1.6kg1.2

x

xI

+⋅=

+=

Evaluate If = I(0.4 m): ( )

( )( )2

22

f

mkg384.0m4.0kg0.8mkg256.0

m4.0

⋅=

+⋅=

= II


∑ == 2frad ωmrTF


Solve for ωf:

mrT

=fω


( )( ) rad/s0.25m0.4kg0.4

N100f ==ω


( ) ( )xxIIL

ωω

== ff

Solve for ( )xω : ( ) ( )xII

x ff ωω =

Substitute numerical values and simplify to obtain:

( ) ( )( )( ) ( ) 2222

2

kg0.8mkg0.256sJ60.9


xxx

+⋅⋅

=+⋅

⋅=ω

Evaluate ( )m8.0ω :

Chapter 10

800

( )( )( )

rad/s5.12m0.8kg0.8mkg0.256

sJ9.60m8.0 22 =+⋅

⋅=ω

Remarks: Note that this is the angular velocity in both instances. Because the disks leave the cylinder with a tangential velocity of Lω2

1 , the angular momentum of the

system remains constant. 96 ••• Picture the Problem The applied torque accelerates the system and increases the tension in the string until it breaks. The work done before the string breaks is the change in the kinetic energy of the system. We can use Newton’s 2nd law to relate the breaking tension to the angular velocity of the system at the instant the string breaks. Once the applied torque is removed, angular momentum is conserved. Express the work done before the string breaks:

2ff2

1f ωIKKW ==∆= (1)

Using the parallel axis theorem and treating the disks as thin disks, express the moment of inertia of the system (see Table 9-1):

( )( )

( ) ( )224122

121

22412

212

121

mcyl

26

2

2

xRmRLM

mxmRMRML

IIxI

+++=

+++=

+=


( ) ( ) ( ) ( )[ ]( ) ( )[ ]

( ) 22

2241

22121

kg0.8mkg384.0

m4.0kg4.02

m4.06m1.6kg1.2

x

x

xI

+⋅=

++

+=

Evaluate If = I(0.4 m): ( )

( )( )2

22

f

mkg512.0m4.0kg0.8mkg384.0

m4.0

⋅=

+⋅=

= II


∑ == 2frad ωmrTF


Solve for ωf:

mrT

=fω

Substitute numerical values and evaluate ωf: ( )( ) rad/s0.25

m0.4kg0.4N100

f ==ω


801

Substitute in equation (1) to express the work done before the string breaks:

2ff2

1 ωIW =


( )( )J160

rad/s25mkg0.512 2221

=

⋅=W


( ) ( )xxIIL

ωω

== ff

Solve for ( )xω : ( ) ( )xIIx ffωω =

Substitute numerical values to obtain: ( ) ( )( )

( )

( ) 22

22

2

kg0.8mkg0.384sJ8.12


x

xx

+⋅⋅

=

+⋅⋅

=ω

*97 ••• Picture the Problem Let the origin of the coordinate system be at the center of the pulley with the upward direction positive. Let λ be the linear density (mass per unit length) of the rope and L1 and L2 the lengths of the hanging parts of the rope. We can use conservation of mechanical energy to find the angular velocity of the pulley when the difference in height between the two ends of the rope is 7.2 m. (a) Apply conservation of energy to relate the final kinetic energy of the system to the change in potential energy:

0=∆+∆ UK or, because Ki = 0,

0=∆+ UK (1)

Express the change in potential energy of the system: ( ) ( )

( ) ( )[ ]( ) ( )

( ) ( )[ ]22i

21i

22f

21f2

1

22i

21i2

122f

21f2

1

2i2i21

1i1i21

2f2f21

1f1f21

if

LLLLg

gLLgLL

gLLgLLgLLgLL

UUU

+−+−=

+++−=

−−−−−=

−=∆

λ

λλ

λλλλ

Chapter 10

802

Because L1 + L2 = 7.4 m, L2i – L1i = 0.6 m, and L2f – L1f = 7.2 m, we obtain:

L1i = 3.4 m, L2i = 4.0 m, L1f = 0.1 m, and L2f = 7.3 m.

Substitute numerical values and evaluate ∆U:

( )( )( ) ( )[

( ) ( ) ]J75.75

m4m3.4

m7.3m0.1

m/s9.81kg/m0.6

22

22

221

−=−−

+×

−=∆U

Express the kinetic energy of the system when the difference in height between the two ends of the rope is 7.2 m:

( )( ) 22

p21

21

222122

p21

21

2212

p21

ω

ωω

ω

RMM

MRRM

MvIK

+=

+=

+=

Substitute numerical values and simplify: ( )[ ]

( ) 22

22

21

21

mkg1076.02

m2.1kg8.4kg2.2

ω

ωπ

⋅=

⎟⎠⎞

⎜⎝⎛+=K

Substitute in equation (1) and solve for ω:

( ) 0J75.75mkg1076.0 22 =−⋅ ω

and

rad/s5.26mkg1076.0

J75.752 =

⋅=ω

(b) Noting that the moment arm of each portion of the rope is the same, express the total angular momentum of the system:

( )( ) ω

ω

ωω

2rp2

1

2r

2p2

1

2rprp

RMM

RMRM

RMILLL

+=

+=

+=+=

(2)

Letting θ be the angle through which the pulley has turned, express U(θ):

( ) ( ) ( )[ ] gRLRLU λθθθ 22i

21i2

1 ++−−=

Express ∆U and simplify to obtain: ( ) ( )( ) ( )[ ]

( )( ) gRLLgR

gLL

gRLRL

UUUUU

θλλθ

λ

λθθ

θ

2ii122

2i2

21i2

1

22i

21i2

1

if 0

−+−=

++

++−−=

−=−=∆

Assuming that, at t = 0, L1i ≈ L2i: gRU λθ 22−≈∆


803

Substitute for K and ∆U in equation (1) to obtain:

( ) 0mkg1076.0 2222 =−⋅ gR λθω

Solve for ω: 2

22

mkg1076.0 ⋅=

gR λθω


( )( )

( )θ

θπω

1-

2

22

s41.1

mkg1076.0

m/s9.81kg/m0.62

m2.1

=

⋅

⎟⎠⎞

⎜⎝⎛

=

Express ω as the rate of change of θ :

( )θθ 1s41.1 −=dtd

⇒ ( )dtd 1s41.1 −=θθ

Integrate θ from 0 to θ to obtain: ( )t1s41.1ln −=θ

Transform from logarithmic to exponential form to obtain:

( ) ( )tet1s41.1 −

=θ

Differentiate to express ω as a function of time:

( ) ( ) ( )tedtdt

1s41.11s41.1−−==

θω

Substitute for ω in equation (2) to obtain:

( ) ( ) ( )teRMML1s41.112

rp21 s41.1

−−+=


( ) ( )[ ] ( ) ( )[ ] ( ) ( )tt eeL11 s41.12s41.11

2

21 s/mkg303.0s41.1

2m2.1kg4.8kg2.2

−−

⋅=⎟⎠⎞

⎜⎝⎛+= −

π

Chapter 10

804

837

Chapter 11 Gravity Conceptual Problems *1 • (a) False. Kepler’s law of equal areas is a consequence of the fact that the gravitational force acts along the line joining two bodies but is independent of the manner in which the force varies with distance. (b) True. The periods of the planets vary with the three-halves power of their distances from the sun. So the shorter the distance from the sun, the shorter the period of the planet’s motion. 2 • Determine the Concept We can apply Newton’s 2nd law and the law of gravity to the satellite to obtain an expression for its speed as a function of the radius of its orbit.

Apply Newton’s 2nd law to the satellite to obtain: ∑ ==

rvm

rGMmF

2

2radial

where M is the mass of the object the satellite is orbiting and m is the mass of the satellite.

Solve for v to obtain:

rGMv =

Thus the speed of the satellite is independent of its mass and:

correct. is )(c

3 •• Picture the Problem The acceleration due to gravity varies inversely with the square of the distance from the center of the moon. Express the dependence of the acceleration due to the gravity of the moon on the distance from its center:

21r

a' ∝

Express the dependence of the acceleration due to the gravity of the moon at its surface on its radius:

2M

1R

a ∝

Chapter 11

838

Divide the first of these expressions by the second to obtain:

2

2M

rR

aa'

=

Solve for a′: ( )

aaRRa

rRa' 16

12

M

2M

2

2M

4===

and correct. is )(d

4 • Determine the Concept Measurement of G is difficult because masses accessible in the laboratory are very small compared to the mass of the earth. 5 • Determine the Concept The escape speed for a planet is given by RGmv 2e = .

Between ve depends on the square root of M, doubling M increases the escape speed by a factor of 2 and correct. is )( a

6 •• Determine the Concept We can take careful measurements of its position in order to determine whether its trajectory is an ellipse, a hyperbola, or a parabola. If the path is an ellipse, it will return; if its path is hyperbolic or parabolic, it will not return. 7 •• Determine the Concept The gravitational field is proportional to the mass within the sphere of radius r and inversely proportional to the square of r, i.e., proportional to .23 rrr =

*8 • Determine the Concept Let m represent the mass of Mercury, MS the mass of the sun, v the orbital speed of Mercury, and R the mean orbital radius of Mercury. We can use Newton’s 2nd law of motion to relate the gravitational force acting on the Mercury to its orbital speed. Use Newton’s 2nd law to relate the gravitational force acting on Mercury to its orbital speed:

Rvm

RmGMF

2

2S

net ==

Simplify to obtain:

UR

mGMR

mGMmv

21

S21S

212

21

−=

⎟⎠⎞

⎜⎝⎛−−==

Gravity

839

or UK 21−=

9 •• Picture the Problem We can use the definition of the gravitational field to express the ratio of the student’s weight at an elevation of two earth radii to her weight at the surface of the earth. Express the weight of the student at the surface of the earth:

2E

E

RmGMmgw ==

Express the weight of the student at an elevation of two earth radii:

( )2E

E

3RmGMmg'w' ==

Express the ratio of w′ to w: ( )

913

2E

E

2E

E

==

RmGM

RmGM

ww'

and correct. is )(d

10 •• Determine the Concept One such machine would be a balance wheel with weights attached to the rim with half of them shielded using Cavourite. The weights on one side would be pulled down by the force of gravity, while the other side would not, leading to rotation, which can be converted into useful work, in violation of the law of the conservation of energy. Estimation and Approximation 11 • Picture the Problem To approximate the mass of the galaxy we’ll assume the galactic center to be a point mass with the sun in orbit about it and apply Kepler’s 3rd law. Using Kepler’s 3rd law, relate the period of the sun T to its mean distance r from the center of the galaxy:

3

s

galaxy

s

2

3

galaxy

22

44 r

MM

G

MrGM

T

ππ

==

Solve for 2

3

Tr

to obtain:

s

2s

galaxy

s

2s

galaxy

2

3

44GM

MM

M

MM

G

Tr

ππ==

Chapter 11

840

If we measure distances in AU and times in years:

14

s

2

=GMπ

and s

galaxy2

3

MM

Tr

=

Substitute numerical values and evaluate Mgalaxy/Ms:

( )11

26

344

s

galaxy

1008.1y10250LY

AU106.3LY103

×=

×

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×××

=M

M

or

s11

galaxy 1008.1 MM ×=

*12 ••• Picture the Problem We can use Kepler’s 3rd law to find the size of the semi-major axis of the planet’s orbit and the conservation of momentum to find its mass. (a) Using Kepler’s 3rd law, relate the period of this planet T to the length r of its semi-major axis:

3

s

Draconis Iota

s

2

3

s

Draconis Iota

s

2

3

Draconis Iota

22

4

4

4

r

MM

GM

r

MMG

M

rGM

T

π

π

π

=

=

=

If we measure time in years, distances in AU, and masses in terms of the mass of the sun:

14

s

2

=MGπ

and 3

s

Draconis Iota

2 1 r

MMT =

Solve for r to obtain: 3

2

s

Draconis Iota TM

Mr =

Substitute numerical values and evaluate r: ( ) AU33.1y5.105.1

32

s

s =⎟⎟⎠

⎞⎜⎜⎝

⎛=

MMr

(b) Apply conservation of momentum to the planet (mass m and speed v) and the star (mass MIota

Draconis and speed V) to obtain:

VMmv Draconis Iota=

Gravity

841

Solve for m to obtain: v

VMm Draconis Iota=

Use its definition to find the speed of the orbiting planet:

m/s1065.2h

s3600d

h24y

d365.25y1.50

AUm101.5AU1.332

2

4

11

×=

×××

⎟⎟⎠

⎞⎜⎜⎝

⎛ ××

=

=∆∆

=

π

Tr

tdv π


( )( )( )

kg1034.2kg1099.105.10112.0

05.10112.00112.0

m/s102.65m/s296

28

30sun

Draconis Iota

4Draconis Iota

×=

×=

==

⎟⎟⎠

⎞⎜⎜⎝

⎛×

=

MM

Mm

Express m in terms of the mass MJ of Jupiter:

3.21kg1090.1kg1034.2

27

28

J

=××

=Mm

or

J3.12 Mm =

Remarks: A more sophisticated analysis, using the eccentricity of the orbit, leads to a lower bound of 8.7 Jovian masses. (Only a lower bound can be established, as the plane of the orbit is not known.) 13 ••• Picture the Problem We can apply Newton’s law of gravity to estimate the maximum angular velocity which the sun can have if it is to stay together and use the definition of angular momentum to find the orbital angular momenta of Jupiter and Saturn. In part (c) we can relate the final angular velocity of the sun to its initial angular velocity, its moment of inertia, and the orbital angular momenta of Jupiter and Saturn. (a) Gravity must supply the centripetal force which keeps an element of the sun’s mass m rotating around it. Letting the radius of the sun be R, apply Newton’s law of gravity to an element of mass m to obtain:

22

RGMmRm <ω

or

ω2R <GMR2

where we’ve used the inequality because we’re estimating the maximum angular velocity which the sun can have if it is to stay together.

Chapter 11

842

Solve for ω: 3R

GM<ω


( )( )( ) rad/s1028.6

m106.96kg1099.1/kgmN10673.6 4

38

302211−

−

×=×

×⋅×<ω

Calculate the period of this motion from its angular velocity:

h78.2s3600

h1s1000.1

rad/s1028.622

4

4

=××=

×== −

πωπT

(b) Express the orbital angular momenta of Jupiter and Saturn:

JJJJ vrmL = and SSSS vrmL =

Express the orbital speeds of Jupiter and Saturn in terms of their periods and distances from the sun:

J

JJ

2T

rv π= and

S

SS

2T

rv π=


J

2JJ

J2

TrmL π

= and S

2SS

S2

TrmL π

=

Substitute numerical values and evaluate LJ and LS:

( ) ( )( )( )

/smkg1093.1

hs3600

dh24

yd365.25y9.11

m10778kg1098.531823182

243

2924

J

2JE

J

⋅×=

×××

××==

ππT

rML

and ( ) ( )( )( )

/smkg1085.7

hs3600

dh24

yd365.25y5.29

m101430kg1098.51.9521.952

242

2924

S

2SE

S

⋅×=

×××

××==

ππT

rML

Express the angular momentum of the sun as a fraction of the sum of the angular momenta of Jupiter and Saturn: ( )

%703.0

/smkg1085.73.19/smkg1091.1

242

241SJ

sun

=

⋅×+⋅×

=

+=

LLLf

Gravity

843

(c) Relate the final angular momentum of the sun to its initial angular momentum and the angular momenta of Jupiter and Saturn:

SJif LLLL ++= or

SJisunfsun LLII ++= ωω

Solve for ωf to obtain:

sun

SJif I

LL ++=ωω

Substitute for ωI and Isun:

2sunsun

SJ

sunf 059.0

2RM

LLT

++=

πω


( )( )( )

rad/s1080.4

m1096.6kg1099.1059.0/smkg1085.73.19

hs3600

dh24d03

2

4

2830

242

f

−×=

××

⋅×++

××=

πω

Note that this result is about 76% of the maximum possible rotation allowed by gravity that we calculated in part (a). Kepler’s Laws 14 • Picture the Problem We can use the relationship between the semi-major axis and the distances of closest approach and greatest separation, together with Kepler’s 3rd law, to find the greatest separation of Alex-Casey from the sun. Letting x represent the greatest distance from the sun, express the relationship between x, the distance of closest approach, and its semi-major axis R:

2AU1.0+

=xR


AU1.02 −= Rx (1)

Apply Kepler’s 3rd law, with the period T measured in years and R in AU to obtain:

32 RT =

Solve for R: 3 2TR =

Substitute numerical values and evaluate R:

( ) AU3.25y4.1273 2 ==R

Substitute in equation (1) and evaluate R:

( ) AU5.50AU1.0AU3.252 =−=x

Chapter 11

844

15 • Picture the Problem We can use Kepler’s 3rd law to relate the period of Uranus to its mean distance from the sun.

Using Kepler’s 3rd law, relate the period of Uranus to its mean distance from the sun:

32 CrT =

where 3219-

s

2

/ms102.9734×==

GMC π

.

Solve for T: 3CrT =

Substitute numerical values and evaluate T:

( ) ( )y0.84

d365.25y1

h24d1

s3600h1s01651.2

m1087.2/ms10973.2

9

3123219

=××××=

××= −T

16 • Picture the Problem We can use Kepler’s 3rd law to relate the period of Hektor to its mean distance from the sun.

Using Kepler’s 3rd law, relate the period of Hektor to its mean distance from the sun:

32 CrT =

where 3219-

s

2

/ms102.9734×==

GMC π

.

Solve for T: 3CrT =


( )

y8.11d365.25

y1h24

d1s3600

h1s01713.3

AUm101.50AU16.5/ms10973.2

8

3113219

=××××=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×××= −T

17 •• Picture the Problem Kepler’s 3rd law relates the period of Icarus to the length of its semimajor axis. The aphelion distance ra is related to the perihelion distance rp and the semimajor axis by .2pa arr =+

Gravity

845

(a) Using Kepler’s 3rd law, relate the period of Icarus to the length of its semimajor axis:

32 CaT =

where 3219

s

2

/ms102.9734 −×==GM

C π.

Solve for a:

32

CTa =

Substitute numerical values and evaluate a:

m1059.1

/ms10973.2h

s3600d

h24y

d365.25.1y1

11

3

3219

2

×=

×

⎟⎟⎠

⎞⎜⎜⎝

⎛×××

= −a

(b) Use the definition of the eccentricity of an ellipse to determine the perihelion distance of Icarus:

( )( )( )

m1071.2

83.01m1059.1

1

10

11

p

×=

−×=

−= ear

Express the relationship between rp and ra for an ellipse:

arr 2pa =+

Solve for and evaluate ra:

( )m1091.2

m1071.2m1059.12

2

11

1011

pa

×=

×−×=

−= rar

18 •• Picture the Problem The Hohmann transfer orbit is shown in the diagram. We can apply Kepler’s 3rd law to relate the time-in-orbit to the period of the spacecraft in its Hohmann Earth-to-Mars orbit. The period of this orbit is, in turn, a function of its semi-major axis which we can find from the average of the lengths of the semi-major axes of the Earth and Mars orbits.

Chapter 11

846

Using Kepler’s 3rd law, relate the period T of the spacecraft to the semi-major axis of its orbit:

32 RT =

Solve for T to obtain:

3RT =

Relate the transit time to the period of this orbit:

321

21

trip RTt ==

Express the semi-major axis of the Hohmann transfer orbit in terms of the mean sun-Mars and sun-Earth distances:

AU1.262

AU1.00AU1.52=

+=R

Substitute numerical values and evaluate ttrip:

( )

d258y1

d365.24y707.0

AU26.1 321

trip

=×=

=t

*19 •• Picture the Problem We can use a property of lines tangent to a circle and radii drawn to the point of contact to show that b = 90°. Once we’ve established that b is a right angle we can use the definition of the sine function to relate the distance from the sun to Venus to the distance from the sun to the earth. (a) The line from earth to Venus' orbit is tangent to the orbit of Venus at the point of maximum extension. Venus will appear closer to the sun in earth’s sky when it passes the line drawn from earth and tangent to its orbit. Hence:

°= 90b

(b) Using trigonometry, relate the distance from the sun to Venus dSV to the angle a:

SE

SVsindda =

Solve for dSV: add sinSESV =

Substitute numerical values and evaluate dSV:

( ) AU731.074sinAU1SV =°=d

Remarks: The correct distance from the sun to Venus is closer to 0.723 AU. 20 •• Picture the Problem Because the gravitational force the Earth exerts on the moon is along the line joining their centers, the net torque acting on the moon is zero and its angular momentum is conserved in its orbit about the Earth. Because energy is also conserved, we can combine these two expressions to solve for either vp or va initially and

Gravity

847

then substitute in the conservation of angular momentum equation to find the other. Letting m be the mass of the moon, apply conservation of angular momentum to the moon at apogee and perigee to obtain:

aapp rmvrmv = or v prp = va ra

Solve for va: p

a

pa v

rr

v = (1)

Apply conservation of energy to the moon-earth system to obtain:

aa

pp r

GMmmvr

GMmmv −=− 22

21

21

or

aa

pp r

GMvr

GMv −=− 22

21

21

Substitute for va to obtain:

ap

a

p

ap

a

p

pp

rGMv

rr

rGMv

rr

rGMv

−⎟⎟⎠

⎞⎜⎜⎝

⎛=

−⎟⎟⎠

⎞⎜⎜⎝

⎛=−

22

22

21

21

21

Solve for vp to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛

+=

appp rrr

GMv1

12

Substitute numerical values and evaluate vp:

( ) ( ) km/s09.1

m10064.4m10576.31

1m10576.3

kg1098.5/kgmN10673.62

8

88

242211

=⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

××

+××⋅×

=−

pv

Substitute numerical values in equation (1) and evaluate va: ( )

km/s959.0

km/s1.09m10064.4m10576.3

8

8

=

××

=av

Newton’s Law of Gravity

*21 •• Picture the Problem We can use Kepler’s 3rd law to find the mass of Jupiter in part (a). In part (b) we can express the centripetal accelerations of Europa and Callisto and compare their ratio to the square of the ratio of their distances from the center of Jupiter

Chapter 11

848

to show that the given data is consistent with an inverse square law for gravity. (a) Assuming a circular orbit, apply Kepler’s 3rd law to the motion of Europa to obtain:

3E

J

22

E4 RGM

T π=

Solve for the mass of Jupiter: 3

E2E

2

J4 RGT

M π=

Substitute numerical values and evaluate MJ: ( )

( )

.kg101.902 of valueacceptedwith theagreement excellent

inresult a ,kg1090.1

hs3600

dh24d3.55

m106.71

/kgmN10673.64

27

27

2

38

2211

2

J

×

×=

⎟⎠⎞

⎜⎝⎛ ××

××

⋅×= −

πM

(b) Express the centripetal acceleration of both of the moons to obtain: 2

2

2

2 42

TR

RT

R

Rv π

π

=⎟⎠⎞

⎜⎝⎛

=

where R and T are the radii and periods of their motion.

Using this result, express the centripetal accelerations of Europa and Callisto:

2E

E2

E4

TRa π

= and 2C

C2

C4

TRa π

=

Substitute numerical values and evaluate aE:

( )( )( )( )[ ]

2

2

82

E

m/s282.0

s/h3600h/d24d55.3m1071.64

=

×=

πa

Substitute numerical values and evaluate aC:

( )( )( )( )[ ]

2

2

82

C

m/s0356.0

s/h3600h/d24d7.16m108.184

=

×=

πa

Evaluate the ratio of these accelerations:

91.7m/s0356.0m/s282.0

2

2

C

E ==aa

Gravity

849

Evaluate the square of the ratio of the distance of Callisto divided by the distance of Europa to obtain:

85.7m1071.6m108.18

2

8

82

E

C =⎟⎟⎠

⎞⎜⎜⎝

⎛××

=⎟⎟⎠

⎞⎜⎜⎝

⎛RR

distance. the of square with theinversely variesforce nalgravitatio that theconclusion the

supportsstrongly nscalculatio last twoour of 1%)(within agreement close The

*22 • Determine the Concept The weight of anything, including astronauts, is the reading of a scale from which the object is suspended or on which it rests. If the scale reads zero, then we say the object is ″weightless.″ The pull of the earth’s gravity, on the other hand, depends on the local value of the acceleration of gravity and we can use Newton’s law of gravity to find this acceleration at the elevation of the shuttle. (a) Apply Newton’s law of gravitation to an astronaut of mass m in a shuttle at a distance h above the surface of the earth:

( )2E

Eshuttle Rh

GmMmg+

=

Solve for gshuttle:

( )2E

Eshuttle Rh

GMg+

=

Substitute numerical values and evaluate gshuttle:

( )( )( )

22

242211

shuttle m/s71.8km6370km400

kg1098.5/kgmN10673.6=

+×⋅×

=−

g

(b)."weightless" be toseem will

astronauts theso on,accelerati same eexactly th earth with theofcenter the towardfalling is shuttle on the everything fall" free"in are they Because

23 • Picture the Problem We can use Kepler’s 3rd law to relate the periods of the moons of Saturn to their mean distances from its center.

(a) Using Kepler’s 3rd law, relate the period of Mimas to its mean distance from the center of Saturn:

3M

S

22

M4 r

GMT π

=

Solve for TM: 3

MS

2

M4 rGM

T π=

Chapter 11

850

(b) Using Kepler’s 3rd law, relate the period of Titan to its mean distance from the center of Saturn:

3T

S

22

T4 r

GMT π

=

Substitute numerical values and evaluate TM:

( )( )( ) s1018.8

/kgmN106726.6kg1069.5m1086.14 4

221126

382

M ×=⋅××

×= −

πT

Solve for rT:

32

S2

TT 4π

GMTr =

Substitute numerical values and evaluate rT:

( ) ( )( ) m1022.14

kg1069.5/kgmN106726.6s1038.1 932

26221126

T ×=×⋅××

=−

πr

24 • Picture the Problem We can use Kepler’s 3rd law to relate the period of the moon to the mass of the earth and the mean earth-moon distance.

(a) Using Kepler’s 3rd law, relate the period of the moon to its mean orbital radius:

3m

E

22

m4 r

GMT π

=

Solve for ME: 3

m2m

2

E4 rGT

M π=

Substitute numerical values and evaluate ME:

( )( )

kg1002.6

hs3600

dh24d3.27/kgmN106.6726

m103.844 242

2211

382

E ×=

⎟⎠⎞

⎜⎝⎛ ××⋅×

×=

−

πM

Remarks: This analysis neglects the mass of the moon; consequently the mass calculated here is slightly too great. 25 • Picture the Problem We can use Kepler’s 3rd law to relate the period of the earth to the mass of the sun and the mean earth-sun distance.

Gravity

851

(a) Using Kepler’s 3rd law, relate the period of the earth to its mean orbital radius:

3E

S

22

E4 r

GMT π

=

Solve for MS: 3

E2E

2

S4 rGT

M π=

Substitute numerical values and evaluate MS:

( )( )

kg1099.1

hs3600

dh24

yd365.25y1/kgmN106.6726

m10496.14

30

22211

3112

S

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛×××⋅×

×=

−

πM

*26 • Picture the Problem We can relate the acceleration of an object at any elevation to its acceleration at the surface of the earth through the law of gravity and Newton’s 2nd law of motion.

Letting a represent the acceleration due to gravity at this altitude (RE) and m the mass of the object, apply Newton’s 2nd law and the law of gravity to obtain:

( )∑ == maR

GmMF 2E

Eradial 2

and

( )2E

E

2RGMa = (1)

Apply Newton’s 2nd law to the same object when it is at the surface of the earth:

∑ == mgR

GmMF 2E

Eradial

and

2E

E

RGMg = (2)

Divide equation (1) by equation (2) and solve for a: 2

E

2E

4RR

ga=

and ( ) 22

41

41 m/s2.45m/s9.81 === ga

27 • Picture the Problem Your weight is the local gravitational force exerted on you. We can use the definition of density to relate the mass of the planet to the mass of earth and the

Chapter 11

852

law of gravity to relate your weight on the planet to your weight on earth.

Using the definition of density, relate the mass of the earth to its radius:

3E3

4EE RVM πρρ ==

Relate the mass of the planet to its radius: ( )3E3

4

3P3

4PP

10R

RVM

πρ

πρρ

=

==

Divide the second of these equations by the first to express MP in terms of ME:

( )3E3

4

3E3

4

E

P 10RR

MM

πρπρ

ρ=

and

E3

P 10 MM =

Letting w′ represent your weight on the planet, use the law of gravity to relate w′ to your weight on earth:

( )( )

wR

GmMR

MGmR

GmMw'

1010

1010

2E

E

2E

E3

2P

P

==

==

where w is your weight on earth. 28 • Picture the Problem We can relate the acceleration due to gravity of a test object at the surface of the new planet to the acceleration due to gravity at the surface of the earth through use of the law of gravity and Newton’s 2nd law of motion.

Letting a represent the acceleration due to gravity at the surface of this new planet and m the mass of a test object, apply Newton’s 2nd law and the law of gravity to obtain:

( )∑ == maR

GmMF 2

E21

Eradial

and

( )2E21

E

RGM

a =

Simplify this expression to obtain: 2

2E

E m/s2.3944 === gR

GMa

29 • Picture the Problem We can use conservation of angular momentum to relate the planet’s speeds at aphelion and perihelion.

Using conservation of angular pa LL =

Gravity

853

momentum, relate the angular momenta of the planet at aphelion and perihelion:

or aapp rmvrmv =

Solve for the planet’s speed at aphelion:

a

ppa r

rvv =

Substitute numerical values and evaluate va:

( )( )

m/s1027.2

m102.2m101.0m/s105

4

15

154

a

×=

×××

=v

30 • Picture the Problem We can use Newton’s law of gravity to express the gravitational force acting on an object at the surface of the neutron star in terms of the weight of the object. We can then simplify this expression be dividing out the mass of the object … leaving an expression for the acceleration due to gravity at the surface of the neutron star. Apply Newton’s law of gravity to an object of mass m at the surface of the neutron star to obtain:

mgR

mGM=2

StarNeutron

StarNeutron

where g represents the acceleration due to gravity at the surface of the neutron star.

Solve for g and substitute for the mass of the neutron star:

( )2

StarNeutron

sun2

StarNeutron

StarNeutron 60.1R

MGR

GMg ==

Substitute numerical values and evaluate g:

( )( )( )

2122

302211

m/s1093.1km10.5

kg1099.1/kgmN10673.660.1×=

×⋅×=

−

g

*31 •• Picture the Problem We can use conservation of angular momentum to relate the asteroid’s aphelion and perihelion distances.

Using conservation of angular momentum, relate the angular momenta of the asteroid at aphelion and perihelion:

pa LL =

or aapp rmvrmv =

Solve for and evaluate the ratio of the asteroid’s aphelion and perihelion distances:

43.1km/s14km/s20

a

p

p

a ===vv

rr

Chapter 11

854

32 •• Picture the Problem We’ll use the law of gravity to find the gravitational force acting on the satellite. The application of Newton’s 2nd law will lead us to the speed of the satellite and its period can be found from its definition.

(a) Letting m represent the mass of the satellite and h its elevation, use the law of gravity to express the gravitational force acting on it:

( ) ( )

2

E

2E

2E

2E

Eg

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

+=

+=

Rh

mghRgmR

hRGmMF

Substitute numerical values and evaluate Fg:

( )( )

N6.37

m106.37m1051

N/kg9.81kg300

12

6

72

E

g

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×+

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

Rh

mgF

(b) Using Newton’s 2nd law, relate the gravitational force acting on the satellite to its centripetal acceleration:

rvmF

2

g =

Solve for v:

mrF

v g=


( )( )

km/s2.66

kg300m105m106.37N37.6 76

=

×+×=v

(c) Express the period of the satellite:

vrT π2

=


( )

h9.36s3600

h1s1033.1

m/s102.66m105m106.372

5

3

76

=××=

××+×

=πT

Gravity

855

*33 •• Picture the Problem We can determine the maximum range at which an object with a given mass can be detected by substituting the equation for the gravitational field in the expression for the resolution of the meter and solving for the distance. Differentiating g(r) with respect to r, separating variables to obtain dg/g, and approximating ∆r with dr will allow us to determine the vertical change in the position of the gravity meter in the earth’s gravitational field is detectable.

(a) Express the gravitational field of the earth: 2

E

EE R

GMg =

Express the gravitational field due to the mass m (assumed to be a point mass) of your friend and relate it to the resolution of the meter:

( ) 2E

E11E

112 1010

RGMg

rGmrg −− ===

Solve for r:

E

11

E10

MmRr =

Substitute numerical values and evaluate r: ( ) ( )

m37.7

kg105.98kg8010m106.37 24

116

=

××=r

(b) Differentiate g(r) and simplify to obtain:

grr

Gmrr

Gmdrdg 222

23 −=⎟⎠⎞

⎜⎝⎛−=

−=

Separate variables to obtain: 11102 −=−=

rdr

gdg

Approximating dr with ∆r, evaluate ∆r with r = RE:

( )( )

mm0319.0

m1019.3

m1037.6105

61121

=

×=

×−=∆−

−r

34 •• Picture the Problem We can use the law of gravity and Newton’s 2nd law to relate the force exerted on the planet by the star to its orbital speed and the definition of the period to relate it to the radius of the orbit.

Chapter 11

856

Using the law of gravity and Newton’s 2nd law, relate the force exerted on the planet by the star to its centripetal acceleration:

rvm

rKMmF

2

net ==


KMv =2

Express the period of the planet: rKMKM

rv

rT πππ 222===

or rT ∝

*35 •• Picture the Problem We can use the definitions of the gravitational fields at the surfaces of the earth and the moon to express the accelerations due to gravity at these locations in terms of the average densities of the earth and the moon. Expressing the ratio of these accelerations will lead us to the ratio of the densities.

Express the acceleration due to gravity at the surface of the earth in terms of the earth’s average density:

EE34

2E

3E3

4E

2E

EE2E

EE

RGR

RGR

VGR

GMg

πρ

πρρ

=

===

Express the acceleration due to gravity at the surface of the moon in terms of the moon’s average density:

MM34

M RGg πρ=

Divide the second of these equations by the first to obtain: EE

MM

E

M

RR

gg

ρρ

=

Solve for E

M

ρρ

: ME

EM

E

M

RgRg

=ρρ

Substitute numerical values and

evaluate E

M

ρρ

:

( )( )( )( )

605.0

m101.738m/s9.81m106.37m/s1.62

62

62

E

M

=

××

=ρρ

Gravity

857

Measurement of G 36 • Picture the Problem We can use the law of gravity to find the forces of attraction between the two masses and the definition of torque to determine the balancing torque required.

(a) Use the law of gravity to express the force of attraction between the two masses:

221

rmGmF =

Substitute numerical values and evaluate F:

( )( )( )( )

N1085.1m0.06

kg0.01kg10/kgmN106.6726 92

2211−

−

×=⋅×

=F

(b) Use its definition to find the torque exerted by the suspension to balance these forces:

( )( )mN103.70

m0.1N1085.12210

9

⋅×=

×==−

−Frτ

Gravitational and Inertial Mass 37 • Picture the Problem Newton’s 2nd law of motion relates the masses and accelerations of these objects to their common accelerating force. (a) Apply Newton’s 2nd law to the standard object:

11amF =

Apply Newton’s 2nd law to the object of unknown mass:

22amF =

Eliminate F between these two equations and solve for m2:

12

12 m

aam =

Substitute numerical values and evaluate m2:

( ) kg2.27kg1m/s1.1705m/s2.6587

2

2

2 ==m

(b) . of mass theisIt 2minertial

Chapter 11

858

38 • Picture the Problem Newton’s 2nd law of motion relates the weights of these two objects to their masses and the acceleration due to gravity. (a) Apply Newton’s 2nd law to the standard object:

gmwF 11net ==

Apply Newton’s 2nd law to the object of unknown mass:

gmwF 22net ==

Eliminate g between these two equations and solve for m2:

11

22 m

wwm =

Substitute numerical values and evaluate m2:

( ) kg77.5kg1N9.81N56.6

2 ==m

(b) . of mass theisit field, nalgravitatio

searth' theof on effect by the determined isresult thisSince

2

2

mnalgravitatiom

*39 • Picture the Problem Noting that g1 ~ g2 ~ g, let the acceleration of gravity on the first object be g1, and on the second be g2. We can use a constant-acceleration equation to express the difference in the distances fallen by each object and then relate the average distance fallen by the two objects to obtain an expression from which we can approximate the distance they would have to fall before we might measure a difference in their fall distances greater than 1 mm. Express the difference ∆d in the distances fallen by the two objects in time t:

21 ddd −=∆

Express the distances fallen by each of the objects in time t:

212

11 tgd =

and 2

221

2 tgd =


( ) 2212

1222

1212

1 tggtgtgd −=−=∆

Relate the average distance d fallen by the two objects to their time of fall:

221 gtd =

or

gdt 22 =

Gravity

859

Substitute to obtain: g

gdgdgd ∆=∆≈∆

221

Solve for d to obtain:

ggdd∆

∆=

Substitute numerical values and evaluate d: ( )( ) m1010m10 9123 == −d

Gravitational Potential Energy 40 • Picture the Problem Choosing the zero of gravitational potential energy to be at infinite separation yields, as the potential energy of a two-body system in which the objects are separated by a distance r, ( ) rGMmrU −= , where M and m are the masses of the two

bodies. In order for an object to just escape a gravitational field from a particular location, it must have enough kinetic energy so that its total energy is zero. (a) Letting U(∞) = 0, express the gravitational potential energy of the earth-object system:

( )r

mGMrU E−= (1)

Substitute for GME and simplify to obtain:

( ) EE

2E

E

EE mgR

RmgR

RmGMRU −=−=−=

Substitute numerical values and evaluate U(RE):

( ) ( )( )( ) J106.25m106.37kg/N9.81kg100 96E ×−=×−=RU

(b) Evaluate equation (1) with r = 2RE: ( )

E21

E

2E

E

EE 22

2

mgRR

mgRR

mGMRU

−=

−=−=

Substitute numerical values and evaluate U(2RE):

( ) ( )( )( ) J1012.3m106.37kg/N9.81kg1002 9621

E ×−=×−=RU

(c) Express the condition that an object must satisfy in order to escape from the earth’s gravitational

( ) ( ) 022 EEesc =+ RURK

or ( ) 02 E

2esc2

1 =+ RUmv

Chapter 11

860

field from a height RE above its surface: Solve for vesc: ( )

mRUv E

esc22−

=

Substitute numerical values and evaluate vesc:

( ) km/s7.90kg100

J103.122 9

esc =×−−

=v

41 • Picture the Problem In order for an object to just escape a gravitational field from a particular location, an amount of work must be done on it that is equal to its potential energy in its initial position. Express the work needed to remove the point mass from the surface of the sphere to a point a very large distance away:

if UUUW −=∆=

or, because Uf = 0, iUUW −=∆= (1)

Express the initial potential energy of the system:

RGMm

U 0i −=


RGMm

W 0=

42 • Picture the Problem Let the zero of gravitational potential energy be at infinity and let m represent the mass of the spacecraft. We’ll use conservation of energy to relate the initial kinetic and potential energies to the final potential energy of the earth-spacecraft system. Use conservation of energy to relate the initial kinetic and potential energies of the system to its final energy when the spacecraft is one earth radius above the surface of the planet:

0ifif =−+− UUKK

or, because Kf = 0, ( ) ( ) ( ) 02 EEE =−+− RURURK (1)

Gravity

861

Express the potential energy of the spacecraft-and-earth system when the spacecraft is at a distance r from the surface of the earth:

( )r

mGMrU E−=

Substitute in equation (1) to obtain: 02 E

E

E

E221 =+−−

RmGM

RmGMmv

Solve for v:

EE

2E

E

E gRRgR

RGMv ===

Substitute numerical values and evaluate v: ( )( )

km/s7.91

m106.37m/s9.81 62

=

×=v

*43 •• Picture the Problem Let the zero of gravitational potential energy be at infinity and let m represent the mass of the object. We’ll use conservation of energy to relate the initial potential energy of the object-earth system to the final potential and kinetic energies. Use conservation of energy to relate the initial potential energy of the system to its energy as the object is about to strike the earth:

0ifif =−+− UUKK

or, because Ki = 0, ( ) ( ) ( ) 0EEE =+−+ hRURURK (1)

where h is the initial height above the earth’s surface.

Express the potential energy of the object-earth system when the object is at a distance r from the surface of the earth:

( )r

mGMrU E−=

Substitute in equation (1) to obtain: 0E

E

E

E221 =

++−

hRmGM

RmGM

mv

Solve for v:

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=

hRhgR

hRGM

RGMv

EE

E

E

E

E

2

2

Chapter 11

862


( )( )( ) km/s94.6m104m106.37

m104m106.37m/s9.81266

662

=×+×

××=v

44 •• Picture the Problem Let the zero of gravitational potential energy be at infinity, m represent the mass of the object, and h the maximum height reached by the object. We’ll use conservation of energy to relate the initial potential and kinetic energies of the object-earth system to the final potential energy. Use conservation of energy to relate the initial potential energy of the system to its energy as the object is about to strike the earth:

0ifif =−+− UUKK

or, because Kf = 0, ( ) ( ) ( ) 0EEE =+−+ hRURURK (1)

where h is the initial height above the earth’s surface.

Express the potential energy of the object-earth system when the object is at a distance r from the surface of the earth:

( )r

mGMrU E−=

Substitute in equation (1) to obtain: 0E

E

E

E221 =

++−

hRmGM

RmGM

mv

Solve for h:

122

E

E

−=

vgR

Rh

Substitute numerical values and evaluate h:

( )( )( )

km359

1m104

m106.37m/s9.812m106.37

23

62

6

=

−×

××

=h

45 •• Picture the Problem When the point mass is inside the spherical shell, there is no mass between it and the center of the shell. On the other hand, when the point mass is outside the spherical shell we can use the law of gravity to express the force acting on it. In (b) we can derive U(r) from F(r).

Gravity

863

(a) The force exerted by the shell on a point mass m0 when m0 is inside the shell is:

0inside =Fr

The force exerted by the shell on a point mass m0 when m0 is outside the shell is:

rgF ˆ2

00outside r

GMmm −==rr

where r is radially outward from the center of the spherical shell.

(b) Use its definition to express U(r) for r > R:

( )

rGMm

drrGMmdrFrUrr

r

0

20

−=

=−= ∫∫∞

−

∞

When r = R: ( )

RGMmRU 0−=

(c) For r < R, F = 0 and: constant0 =⇒= U

drdU

(d) Because U is continuous, then for r < R:

( ) ( )R

GMmRUrU 0−==

(e) A sketch of U as a function of r/R (with GMm0 = 1) is shown below:

GMm 0 = 1

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0 1 2 3 4

r /R

U

Chapter 11

864

46 • Picture the Problem The escape speed from a planet is related to its mass according to

RGMv 2e = , where M and R represent the mass and radius of the planet,

respectively. Express the escape speed from Saturn:

S

Se.S

2R

GMv = (1)

Express the escape speed from Earth:

E

Ee.E

2R

GMv = (2)

Divide equation (1) by equation (2) to obtain:

E

S

S

E

E

E

S

S

e.E

e.S

2

2

MM

RR

RGMR

GM

vv

⋅==

Substitute numerical values and

evaluate e.E

e.S

vv

:

17.31

2.9547.91

e.E

e.S =×=vv

Solve for and evaluate ve,S: ( )km/s5.35

km/s2.1117.317.3 e.Ee.S

=

== vv

47 • Picture the Problem The escape speed from the moon or the earth is given by

RGMv 2e = , where M and R represent the masses and radii of the moon or the earth.

Express the escape speed from the moon:

mmm

me.S 22 Rg

RGMv == (1)

Express the escape speed from earth:

EEE

Ee.E 22 Rg

RGMv == (2)


EE

mm

EE

mm

e.E

e.m

RgRg

RgRg

vv

==

Gravity

865

Solve for ve,m: e.E

EE

mme.m v

RgRgv =

Substitute numerical values and evaluate ve,m:

( )( )( )km/s38.2

km/s2.11273.0166.0e.m

=

=v

*48 • Picture the Problem We’ll consider a rocket of mass m which is initially on the surface of the earth (mass M and radius R) and compare the kinetic energy needed to get the rocket to its escape velocity with its kinetic energy in a low circular orbit around the earth. We can use conservation of energy to find the escape kinetic energy and Newton’s law of gravity to derive an expression for the low earth-orbit kinetic energy. Apply conservation of energy to relate the initial energy of the rocket to its escape kinetic energy:

0ifif =−+− UUKK

Letting the zero of gravitational potential energy be at infinity we have Uf = Kf = 0 and:

0ii =−− UK or

RGMmUK =−= ie

Apply Newton’s law of gravity to the rocket in orbit at the surface of the earth to obtain:

Rvm

RGMm 2

2 =

Rewrite this equation to express the low-orbit kinetic energy Eo of the rocket: R

GMmmvK2

221

o ==

Express the ratio of Ko to Ke:

212

e

o ==

RGMm

RGMm

KK

⇒ oe 2KK = , as

asserted by Heinlein. 49 •• Picture the Problem Let the zero of gravitational potential energy be at infinity, m represent the mass of the particle, and the subscript E refer to the earth. When the particle is very far from the earth, the gravitational potential energy of the earth-particle system will be zero. We’ll use conservation of energy to relate the initial potential and kinetic energies of the particle-earth system to the final kinetic energy of the particle. Use conservation of energy to relate the initial energy of the system to its energy when the particle is very

0ifif =−+− UUKK

or, because Uf = 0, ( ) ( ) ( ) 0EE =−−∞ RURKK (1)

Chapter 11

866

from the earth:

Substitute in equation (1) to obtain: ( ) 02E

E2e2

1221 =+−∞ R

mGMvmmv

or, because 2EE gRGM = ,

0E2

212

21 =+−∞ mgRmvmv

Solve for v∞: ( )E2e22 gRvv −=∞

Substitute numerical values and evaluate v∞:

( ) ( )( )[ ] km/s4.19m106.37m/s9.81m/s1011.222 6223 =×−×=∞v

50 •• Picture the Problem Let the zero of gravitational potential energy be at infinity, m represent the mass of the particle, and the subscript E refer to the earth. When the particle is very far from the earth, the gravitational potential energy of the earth-particle system will be zero. We’ll use conservation of energy to relate the initial potential and kinetic energies of the particle-earth system to the final kinetic energy of the particle. Use conservation of energy to relate the initial energy of the system to its energy when the particle is very far away:

0ifif =−+− UUKK

or, because Uf = 0, ( ) ( ) ( ) 0EE =−−∞ RURKK (1)


E

E2i2

1221 =+−∞ R

mGMmvmv

or, because 2EE gRGM = ,

0E2i2

1221 =+−∞ mgRmvmv

Solve for vi: E

2i 2gRvv += ∞

Substitute numerical values and evaluate vi:

( ) ( )( ) km/s8.15m106.37m/s9.812m/s1011.2 6223i =×+×=v

51 •• Picture the Problem We can use the definition of kinetic energy to find the energy necessary to launch a 1-kg object from the earth at escape speed.

Gravity

867

(a) Using the definition of kinetic energy, find the energy required to launch a 1-kg object from the surface of the earth at escape speed:

( )( )MJ62.7

m/s1011.2kg1 2321

2e2

1

=

×=

= mvK

(b) Using the conversion factor 1 kW⋅h = 3.6 MJ, convert 62.7 MJ to kW⋅h: hkW4.17

MJ3.6hkW1MJ7.62

⋅=

⋅×=K

(c) Express the cost of this project in terms of the mass of the astronaut:

masskg

energyrequiredrateCost ××=

Substitute numerical values and find the cost:

( )

$139

kg80kg

hkW17.4h kW

$0.10Cost

=

⋅×

⋅=

52 •• Picture the Problem Let m represent the mass of the body that is projected vertically from the surface of the earth. We’ll begin by using conservation of energy under the assumption that the gravitational field is constant to determine H ′. We’ll apply conservation of energy a second time, with the zero of gravitational potential energy at infinity, to express H. Finally, we’ll solve these two equations simultaneously to express H in terms of H ′. Assuming the gravitational field to be constant and letting the zero of potential energy be at the surface of the earth, apply conservation of energy to relate the initial kinetic energy and the final potential energy of the object-earth system:

0ifif =−+− UUKK

or, because Kf = Ui = 0, 0fi =+− UK

Substitute for Ki and Uf and solve for H ′:

0221 =+− mgH'mv

and

gvH'2

2

= (1)

Letting the zero of gravitational potential energy be at infinity, use conservation of energy to relate the initial kinetic energy and the final

0ifif =−+− UUKK

or, because Kf = 0, 0ifi =−+− UUK

Chapter 11

868

potential energy of the object-earth system: Substitute to obtain: 0

EE

221 =+

+−−

RGMm

HRGMmmv

or

0E

2E

E

2E2

21 =+

+−−

RgR

HRgRv

Solve for v2:

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=

HRHgR

HRRgRv

EE

EE

2E

2

2

112


⎟⎟⎠

⎞⎜⎜⎝

⎛+

=HR

HRH'E

E

Solve for H:

H'RH'RH−

=E

E

Orbits 53 •• Picture the Problem We can use its definition to express the period of the spacecraft’s motion and apply Newton’s 2nd law to the spacecraft to determine its orbital velocity. We can then use this orbital velocity to calculate the kinetic energy of the spacecraft. We can relate the spacecraft’s angular momentum to its kinetic energy and moment of inertia. (a) Express the period of the spacecraft’s orbit about the earth:

( )vR

vR

vRT EE 6322 πππ

===

where v is the orbital speed of the spacecraft.

Use Newton’s 2nd law to relate the gravitational force acting on the spacecraft to its orbital speed:

( ) E

2

2E

Eradial 33 R

vmR

mGMF ==


3EgRv =

Gravity

869

Substitute for v in our expression for T to obtain: g

RT E36 π=


h7.31s3600

h1s102.631

m/s9.81m106.37π36

4

2

6

=××=

×=T

(b) Using its definition, express the spacecraft’s kinetic energy:

( )E31

212

21 gRmmvK ==


( )( )( )GJ 1.04

m106.37m/s9.81kg100 6261

=

×=K

(c) Express the kinetic energy of the spacecraft in terms of its angular momentum:

ILK2

2

=

Solve for L: IKL 2=

Express the moment of inertia of the spacecraft with respect to an axis through the center of the earth:

( )2E

2E

9

3

mR

RmI

=

=

Substitute and solve for L: mKRKmRL 2318 E2E ==


( ) ( )( ) sJ108.72J101.04kg1002m106.373 1296 ⋅×=××=L

*54 • Picture the Problem Let the origin of our coordinate system be at the center of the earth and let the positive x direction be toward the moon. We can apply the definition of center of mass to find the center of mass of the earth-moon system and find the ″orbital″ speed of the earth using xcm as the radius of its motion and the period of the moon as the period of this motion of the earth. (a) Using its definition, express the x coordinate of the center of mass of the earth-moon system:

moonE

moonmoonEEcm mM

xmxMx++

=

Chapter 11

870

Substitute numerical values and evaluate xcm:

( ) ( )( ) m1064.4kg1036.7kg1098.5

m1082.3kg1036.70 62224

822E

cm ×=×+×

××+=

Mx

Note that, because the radius of the earth is 6.37×106 m, the center of mass is actually located about 1700 km below the surface of the earth. (b) Express the ″orbital″ speed of the earth in terms of the radius of its circular orbit and its period of rotation:

Txv cm2π

=


( ) m/s4.12

hs3600

dh24d3.27

m1064.42 6

=××

×=

πv

55 •• Picture the Problem We can express the energy difference between these two orbits in terms of the total energy of a satellite at each elevation. The application of Newton’s 2nd law to the force acting on a satellite will allow us to express the total energy of each satellite as function of its mass, the radius of the earth, and its orbital radius. Express the energy difference: 1000geo EEE −=∆ (1)

Express the total energy of an orbiting satellite:

RmGM

mv

UKE

E221

tot

−=

+= (2)

where R is the orbital radius.

Apply Newton’s 2nd law to a satellite to relate the gravitational force to the orbital speed:

Rvm

RmGMF

2

2E

radial ==

or

Rv

RgR 2

2

2E =

Simplify and solve for v2:

RgRv

2E2 =


RmgR

RmgR

RgRmE

2

2E

2E

2E

21

tot −=−=

Gravity

871


⎟⎟⎠

⎞⎜⎜⎝

⎛−=

+−=∆

geo1000

2E

1000

2E

geo

2E

112

22

RRmgR

RmgR

RmgRE


( )( )( ) GJ1.11m104.22

1m1037.7

1m106.37kg/N9.81kg500 76

2621 =⎟⎟

⎠

⎞⎜⎜⎝

⎛×

−×

×=∆E

56 •• Picture the Problem We can use Kepler’s 3rd law to relate the periods of the moon and Earth, in their orbits about the earth and the sun, to their mean distances from the objects about which they are in orbit. We can solve these equations for the masses of the sun and the earth and then divide one by the other to establish a value for the ratio of the mass of the sun to the mass of the earth. Using Kepler’s 3rd law, relate the period of the moon to its mean distance from the earth:

3m

E

22

m4 r

GMT π

= (1)

where rm is the distance between the centers of the earth and the moon.

Using Kepler’s 3rd law, relate the period of the earth to its mean distance from the sun:

3E

s

22

E4 rGM

T π= (2)

where rE is the distance between the centers of the earth and the sun.

Solve equation (1) for ME: 3

m2m

2

E4 rGT

M π= (3)

Solve equation (2) for Ms: 3

E2E

2

s4 rGT

M π= (4)

Divide equation (4) by equation (3) and simplify to obtain:

2

E

m

3

m

E

E

s⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

TT

rr

MM

Substitute numerical values and evaluate Ms/ME:

5

23

8

11

E

s

1038.3

d24.365d3.27

m1082.3m105.1

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛××

=MM

Chapter 11

872

Express the difference between this value and the measured value of 3.33×105:

%50.1103.33

103.33103.38diff% 5

55

=×

×−×=

The Gravitational Field 57 • Picture the Problem The gravitational field at any point is defined by .mFg

rr=

Using its definition, express the gravitational field at a point in space:

( ) ( )iiFg ˆN/kg4kg3

ˆN12===

m

rr

*58 • Picture the Problem The gravitational field at any point is defined by .mFg

rr=

Using its definition, express the gravitational field at a point in space:

mFgr

r=

Solve for Fr

and substitute for m and gr

to obtain: ( )( )( )j

j

gF

ˆ10

ˆN/kg102.5kg004.08

6

N

m

−

−

=

×=

=rr

59 •• Picture the Problem We can use the definition of the gravitational field due to a point mass to find the x and y components of the field at the origin and then add these components to find the resultant field. We can find the magnitude of the field from its components using the Pythagorean theorem. (a) Express the gravitational field due to the point mass at x = L:

ig ˆ2L

Gmx =r

Express the gravitational field due to the point mass at y = L:

jg ˆ2L

Gmy =r

Add the two fields to obtain:

jiggg ˆˆ22 L

GmL

Gmyx +=+=rrr

Gravity

873

(b) Find the magnitude of :gr

2

2222

2L

GmL

GmL

Gmgg yx

=

+=+=gr

60 •• Picture the Problem We can find the net force acting on m by superposition of the forces due to each of the objects arrayed on the circular arc. Once we have expressed the net force, we can find the gravitational field at the center of curvature from its definition. (a) Express the net force acting on m:

jiF ˆˆyx FF +=

r (1)

Express Fx:

0

45cos

45cos

2

222

=

°−

°+−=

RGMm

RGMm

RGMm

RGMmFx

Express Fy:

( )145sin2

45sin

45sin

2

2

22

+°=

°+

°+=

RGMm

RGMm

RGMm

RGMmFy

Substitute numerical values and evaluate Fy:

( )( )

( )( )( )N1067.9

145sin2kg2kg3m1.0

kg/mN10673.6

8

2

2211

−

−

×=

+°×

⋅×=yF

Substitute in equation (1) to obtain: ( )jiF ˆN1067.9ˆ0 8−×+=r

(b) Using its definition, express g

rat

the center of curvature of the arc: ( )

( )j

jiFg

ˆN/kg1083.4

kg2

ˆN1067.9ˆ0

8

8

−

−

×=

×+==

m

rr

Chapter 11

874

61 •• Picture the Problem The configuration of point masses is shown to the right. The gravitational field at any point can be found by superimposing the fields due to each of the point masses.

(a) Express the gravitational field at x = 2 m as the sum of the fields due to the point masses m1 and m2:

21 gggrrr

+= (1)

Express 1gr

and :2gr

ig ˆ21

11 x

Gm−=

r and ig ˆ

22

22 x

Gm=

r


( )

( )i

ii

iig

ˆ

ˆ2

ˆ

ˆˆ

241

121

21

221

1

22

221

1

mmxG

xGm

xGm

xGm

xGm

−−=

+−=

+−=r

Substitute numerical values and evaluate g

r:

( )( )[ ]

( ) i

i

g

ˆN/kg1067.1

ˆkg4kg2

m2/kgmN106726.6

11

41

2

2211

−

−

×−=

−×

⋅×−=

r

(b) Express 1gr and 2gr : ig ˆ

21

11 x

Gm−=

r and ig ˆ

22

22 x

Gm−=

r


( )

( )i

ii

iig

ˆ

ˆˆ2

ˆˆ

2141

22

22

22

2

1

22

221

1

mmxG

xGm

xGm

xGm

xGm

+−=

−−=

−−=r

Gravity

875

Substitute numerical values and evaluate g

r: ( )

( )[ ]( )i

i

g

ˆN/kg1034.8

ˆkg4kg2

m6/kgmN106726.6

12

41

2

2211

−

−

×−=

+×

⋅×−=

r

(c) Express the condition that g

r= 0:

( )0

6 22

21 =

−−

xGm

xGm

or

( )0

642

22 =−

−xx

Express this quadratic equation in standard form:

036122 =−+ xx , where x is in meters.

Solve the equation to obtain:

m5.14andm48.2 −== xx

From the diagram it is clear that the physically meaningful root is the positive one at:

m48.2=x

62 •• Picture the Problem To show that the maximum value of xg for the field of Example

11-7 occurs at the points ,ax 2±= we can differentiate gx with respect to x and set

the derivative equal to zero. From Example 11-7:

( ) 2/322

2ax

GMxg x+

−=

Differentiate gx with respect to x and set the derivative equal to zero to find extreme values:

( ) ( )[ ] extrema.for 032 2/52222/322 =+−+−=−− axxaxGM

dxdgx


2ax ±=

Chapter 11

876

Remarks: To establish that this value for x corresponds to a relative maximum, we need to either evaluate the second derivative of gx at x = ± a/ 2 or examine the graph of xg at x = ± a/ 2 for concavity downward.

63 •• Picture the Problem We can find the mass of the rod by integrating dm over its length. The gravitational field at x0 > L can be found by integrating g

rd at x0 over the length of the

rod. (a) Express the total mass of the stick:

221

00

CLxdxCdxMLL

=== ∫∫ λ

(b) Express the gravitational field due to an element of the stick of mass dm:

( ) ( )

( )i

iig

ˆ

ˆˆ

20

20

20

xxGCxdx

xxdxG

xxGdmd

−−=

−−=

−−=

λr

Integrate this expression over the length of the stick to obtain: ( )

i

ig

ˆln2

ˆ

00

02

02

0

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

−−= ∫

LxL

Lxx

LGM

xxxdxGC

Lr

64 ••• Picture the Problem Choose a mass element dm of the rod of thickness dx at a distance x from the origin. All such elements produce a gravitational field at a point P located a distance Lx 2

10 > from the origin. We can calculate the total field by integrating the

magnitude of the field produced by dm from x = −L/2 to x = +L/2. (a) Express the gravitational field at P due to the element dm:

ig ˆ2r

Gdmd x −=r

Relate dm to dx: dx

LMdm =

Express the distance r between dm and point P in terms of x and x0:

xxr −= 0

Substitute these results to express xdgr in terms of x and x0: ( )

ig ˆ2

0 ⎭⎬⎫

⎩⎨⎧

−−= dx

xxLGMd x

r

Gravity

877

(b) Integrate to find the total field:

( )

i

i

ig

ˆ

ˆ1

ˆ

2412

0

2/

2/0

2/

2/2

0

LxGM

xxLGM

xxdx

LGM

L

L

L

Lx

−−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−=

−−=

−

−∫

r

(c) Use the definition of g

rto

express :Fr

igF ˆ

2412

0

00 Lx

GMmm−

−==rr

(d) Factor 2

0x from the denominator of our expression for xgr to obtain:

ig ˆ

41 2

0

220 ⎟⎟

⎠

⎞⎜⎜⎝

⎛−

−=

xLx

GMxr

For x0 >> L the second term in parentheses is very small and:

ig ˆ20x

GMx −≈r

which is the gravitational field of a point mass M located at the origin.

gr due to Spherical Objects

65 • Picture the Problem The gravitational field inside a spherical shell is zero and the field at the surface of and outside the shell is given by 2rGMg = .

(a) Because 0.5 m < R: 0=g

(b) Because 1.9 m < R: 0=g

(c) Because 2.5 m > R:

( ) ( )( )

N/kg1020.3

m2.5kg300/kgmN106.6726

9

2

2211

2

−

−

×=

⋅×=

=r

GMg

66 •

Chapter 11

878

Determine the Concept The gravitational attraction is zero. The gravitational field inside the 2 m shell due to that shell is zero; therefore, it exerts no force on the 1 m shell, and, by Newton’s 3rd law, that shell exerts no force on the larger shell. *67 • Picture the Problem The gravitational field and acceleration of gravity at the surface of a sphere given by ,2RGMg = where R is the radius of the sphere and M is its mass.

Express the acceleration of gravity on the surface of S1:

21 RGMg =

Express the acceleration of gravity on the surface of S2: 22 R

GMg =

Divide the second of these equations by the first to obtain: 1

2

2

1

2 ==

RGMR

GM

gg

or 21 gg =

68 •• Picture the Problem The gravitational field and acceleration of gravity at the surface of a sphere given by ,2RGMg = where R is the radius of the sphere and M is its mass.

Express the acceleration of gravity on the surface of S1:

21

1 RGMg =

Express the acceleration of gravity on the surface of S2: 2

22 R

GMg =

Divide the second of these equations by the first to obtain:

22

21

21

22

1

2

RR

RGMR

GM

gg

==

Solve for g2:

122

21

2 gRRg =

Remarks: The accelerations depend only on the masses and radii because the points of interest are outside spherically symmetric distributions of mass. 69 ••

Gravity

879

Picture the Problem The magnitude of the gravitational force is F = mg where g inside a spherical shell is zero and outside is given by .2rGMg =

(a) At r = 3a, the masses of both spheres contribute to g:

( )( )

( )2

21

221

9

3

aMMGm

aMMGmmgF

+=

+==

(b) At r = 1.9a, g due to M2 = 0:

( ) 21

21

61.39.1 aGmM

aGMmmgF ===

(c) At r = 0.9a, g = 0: 0=F

70 •• Picture the Problem The configuration is shown on the right. The centers of the spheres are indicated by the center-lines. The x coordinates of the mass m for parts (a), (b), and (c) are indicated along the x axis. The magnitude of the gravitational force is F = mg where g inside a spherical shell is zero and outside is given

by 2rGMg = .

(a) Express the force acting on the object whose mass is m:

( )xx ggmF 21 +=

Find g1x at x = 3a: ( ) 2

121

1 93 aGM

aGMg x ==

Find g2x at x = 3a:

( ) 22

22

2 84.48.03 aGM

aaGMg x =−

=


⎟⎠⎞

⎜⎝⎛ +=

⎟⎠⎞

⎜⎝⎛ +=

84.49

84.49

212

22

21

MMaGm

aGM

aGMmF

Chapter 11

880

(b) Find g2x at x = 1.9a: ( ) 2

22

22 21.18.09.1 a

GMaa

GMg x =−

=

Find g1x at x = 1.9a: 01 =xg


22

21.1 aGmMmgF ==

(c) At x = 0.9a, g1x = g2x = 0: 0=F

gr Inside Solid Spheres

*71 •• Picture the Problem The "weight" as measured by a spring scale will be the normal force which the spring scale presses up against you. There are two forces acting on you as you stand at a distance r from the center of the planet: the normal force (FN) and the force of gravity (mg). Because you are in equilibrium under the influence of these forces, your weight (the scale reading or normal force) will be equal to the gravitational force acting on you. We can use Newton’s law of gravity to express this force. (a) Express the force of gravity acting on you when you are a distance r from the center of the earth:

2)(

rmrGMFg = (1)

Using the definition of density, express the density of the earth between you and the center of the earth and the density of the earth as a whole:

( )( )

( )3

34 r

rMrVrM

πρ ==

and

334

E

E

E

RM

VM

πρ ==

Because we’re assuming the earth to of uniform-density and perfectly spherical:

( )3

34

E3

34 R

MrrM

ππ=

or

( )3

E ⎟⎠⎞

⎜⎝⎛=

RrMrM


Rr

RmGM

r

mRrGM

Fg 2E

2

3

E

=⎟⎠⎞

⎜⎝⎛

=

Gravity

881

Apply Newton’s law of gravity to yourself at the surface of the earth to obtain:

2E

RmGMmg =

or

2E

RGMg =

where g is the magnitude of free-fall acceleration at the surface of the earth.

Substitute to obtain: r

RmgFg =

i.e., the force of gravity on you is proportional to your distance from the center of the earth.

(b) Apply Newton’s 2nd law to your body to obtain:

2N ωmr

RrmgF −=−

Solve for your ″effective weight″ (i.e., what a spring scale will measure) FN:

rmR

mgmrrR

mgF ⎟⎠⎞

⎜⎝⎛ −=−= 22

N ωω

Note that this equation tells us that your effective weight increases linearly with distance from the center of the earth. The second term can be interpreted as a "centrifugal force" pushing out, which increases the farther you get from the center of the earth.

(c) We can decide whether the change in mass with distance from the center of the earth or the rotational effect is more important by examining the ratio of the two terms in the expression for your effective weight:

RgT

TR

gRg

mr

rR

mg

2

2

222 42 ππωω=

⎟⎠⎞

⎜⎝⎛

==

Substitute numerical values and evaluate this ratio: ( )

( )291

km63704πh

s3600h24m/s9.81

4 2

22

2

2

=

⎟⎠⎞

⎜⎝⎛ ×

=R

gTπ

effect. rotational thethan important more times291 iscenter thefromaway moveyou

asearth theofcenter theandyou between mass in the change The

Chapter 11

882

72 •• Picture the Problem We can find the loss in weight at this depth by taking the difference between the weight of the student at the surface of the earth and her weight at a depth d = 15 km. To find the gravitational field at depth d, we’ll use its definition and the mass of the earth that is between the bottom of the shaft and the center of the earth. We’ll assume (incorrectly) that the density of the earth is constant. Express the loss in weight: ( )RwRww −=∆ )( E (1)

Express the mass M inside R =RE – d:

( )3E34 dRVM −== ρπρ

Express the mass of the earth: 3E3

4EE RVM ρπρ ==

Divide the first of these equations by the second to obtain:

( ) ( )3E

3E

3E3

4

3E3

4

E RdR

RdR

MM −

=−

=ρπ

ρπ

Solve for M: ( )

3E

3E

E RdRMM −

=

Express the gravitational field at R =RE – d:

( )( ) 2

E2

E

3EE

2 RdRdRGM

RGMg

−−

== (2)

Express the gravitational field at R =RE: 2

E

EE R

GMg = (3)


( )( )

E

E

2E

E

2E

2E

3EE

E RdR

RGM

RdRdRGM

gg −

=−

−

=

Solve for g:

EE

E gR

dRg −=

Express the weight of the student at R =RE – d:

( ) ( )

EE

EE

E

R1 mgd

mgR

dRRmgRw

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−==

Gravity

883


E

EE

EE R

1R

dmgmgdmgw =⎟⎟⎠

⎞⎜⎜⎝

⎛−−=∆

Substitute numerical values and evaluate ∆w:

( )( ) N88.1km6370

km15N800==∆w

73 •• Picture the Problem We can use the hint to find the gravitational field along the x axis. Using the hint, express ( )xg : ( ) spherehollowspheresolid ggxg +=

Substitute for gsolid sphere and ghollow sphere and simplify to obtain:

( )( )

( ) ( )[ ]( )

( ) ⎥⎦⎤

⎢⎣

⎡

−−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−

+=

−+=

2212

30

221

321

34

02

334

0

221

spherehollow2

spheresolid

811

34

RxxRG

RxRG

xRG

RxGM

xGM

xg

πρ

πρπρ

74 ••• Picture the Problem The diagram shows the portion of the solid sphere in which the hollow sphere is embedded. 1gr is the field

due to the solid sphere of radius R and density ρ0 and 2gr is the field due to the

sphere of radius ½R and negative density ρ0 centered at ½R. We can find the resultant field by adding the x and y components of 1gr and 2gr .

Use its definition to express 1gr :

34

34

0

2

30

20

21

rGr

Grr

VGr

GM

πρ

πρρ

=

===gr

Find the x and y components of 1gr :

34cos 0

111Gx

rxggg x

πρθ −=⎟⎠⎞

⎜⎝⎛−=−=

and

Chapter 11

884

34sin 0

111Gy

ryggg y

πρθ −=⎟⎠⎞

⎜⎝⎛−=−=

where the negative signs indicate that the field points inward.

Use its definition to express 2gr

:

34

34

20

22

320

22

202

22

Grr

Grr

VGr

GM

πρ

πρρ

=

===gr

where ( ) 2221

2 yRxr +−=

Express the x and y components of :2g

r ( )

34 2

10

2

21

22RxG

rRx

gg x−

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

πρ

34 0

222

Gyrygg y

πρ=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

Add the x components to obtain the x component of the resultant field: ( )

32

34

34

0

21

00

21

GR

RxGGxggg xxx

πρ

πρπρ

−=

−+−=

+=

where the negative sign indicates that the field points inward.

Add the y components to obtain the y component of the resultant field:

03

43

4 00

21

=+−=

+=

GyGy

ggg yyy

πρπρ

Express g

rin vector form and

evaluate gr

: ijig ˆ3

2ˆˆ 0 ⎟⎠⎞

⎜⎝⎛−=+=

GRGg yxπρr

and

32 0GRπρ

=gr

75 ••• Picture the Problem The gravitational field will exert an inward radial force on the objects in the tunnel. We can relate this force to the angular velocity of the planet by using Newton’s 2nd law of motion.

Gravity

885

Letting r be the distance from the objects to the center of the planet, use Newton’s 2nd law to relate the gravitational force acting on the objects to their angular velocity:

2gnet ωmrFF ==

or 2ωmrmg =


rg

=ω (1)

Use its definition to express g:

34

34

0

2

30

20

2

rGr

Grr

VGr

GMg

πρ

πρρ

=

===

Substitute in equation (1) and simplify:

343

40

0G

r

rGπρ

πρ

ω ==

76 ••• Picture the Problem Because we’re given the mass of the sphere, we can find C by expressing the mass of the sphere in terms of C. We can use its definition to find the gravitational field of the sphere both inside and outside its surface. (a) Express the mass of a differential element of the sphere:

( )drrdVdm 24πρρ ==

Integrate to express the mass of the sphere in terms of C: ( ) CrdrCM ππ 2

m5

0

m504 == ∫

Solve for C:

( )π2m50MC =

Substitute numerical values and evaluate C: ( )

22 kg/m436.6

m50kg1011

==π

C

(b) Use its definition to express the gravitational field of the sphere at a distance from its center greater than its radius:

2rGMg =

Chapter 11

886

(1) For r > 5 m: ( ) ( )

2

8

2

2211

N/kg1075.6

kg1011/kgmN106.6726

r

rg

−

−

×=

⋅×=

Use its definition to express the gravitational field of the sphere at a distance from its center less than its radius:

GCr

drrCG

r

drrCr

Gr

drrGg

r

rr

ππ

πρπ

24

44

20

20

2

20

2

==

==

∫

∫∫

(2) For r < 5 m: ( )

( )N/kg1070.2

kg/m6.436/kgmN106726.62

9

2

2211

−

−

×=

×

⋅×= πg

Remarks: Note that g is continuous at r = 5 m. *77 ••• Picture the Problem We can use conservation of energy to relate the work done by the gravitational field to the speed of the small object as it strikes the bottom of the hole. Because we’re given the mass of the sphere, we can find C by expressing the mass of the sphere in terms of C. We can then use its definition to find the gravitational field of the sphere inside its surface. The work done by the field equals the negative of the change in the potential energy of the system as the small object falls in the hole. Use conservation of energy to relate the work done by the gravitational field to the speed of the small object as it strikes the bottom of the hole:

0if =∆+− UKK

or, because Ki = 0 and W = −∆U, 2

21 mvW =

where v is the speed with which the object strikes the bottom of the hole and W is the work done by the gravitational field.

Solve for v:

mWv 2

= (1)

Express the mass of a differential element of the sphere:

( )drrdVdm 24πρρ ==

Gravity

887

Integrate to express the mass of the sphere in terms of C: ( ) CrdrCM ππ 2

m5

0

m504 == ∫

Solve for and evaluate C:

( ) ( )2

22

kg/m436.6m50

kg1011m50

=

==ππ

MC

Use its definition to express the gravitational field of the sphere at a distance from its center less than its radius:

GCr

drrCG

r

drrCr

Gr

drrGg

r

rr

ππ

πρπ

24

44

20

20

2

20

2

==

==

∫

∫∫

Express the work done on the small object by the gravitational force acting on it:

( )mgmgdrW m2m3

m5

=−= ∫


( ) ( ) ( ) GCm

GCmv ππ m82m22==


( ) ( )( ) mm/s104.0kg/m436.6/kgmN106726.6m8 22211 =⋅×= −πv

78 ••• Picture the Problem The spherical deposit of heavy metals will increase the gravitational field at the surface of the earth. We can express this increase in terms of the difference in densities of the deposit and the earth and then form the quotient ∆g/g. Express ∆g due to the spherical deposit:

2rMGg ∆

=∆ (1)

Express the mass of the spherical deposit:

( ) 3343

34 RRVM ρππρρ ∆=∆=∆=

Substitute in equation (1): 2

334

rRG

gρπ ∆

=∆

Chapter 11

888

Express ∆g/g:

2

3342

334

grRG

gr

RG

gg ρπ

ρπ∆

=

∆

=∆

Substitute numerical values and evaluate ∆g/g:

( ) ( ) ( )( )( )

52

33221134

1056.3m2000N/kg81.9

m1000kg/m5000/kgmN106726.6 −−

×=⋅×

=∆ πgg

*79 ••• Picture the Problem The force of attraction of the small sphere of mass m to the lead sphere is the sum of the forces due to the solid sphere ( SF

r) and the cavities ( CF

r) of

negative mass. (a) Express the force of attraction: CS FFF

rrr+= (1)

Use the law of gravity to express the force due to the solid sphere:

iF ˆ2S d

GMm−=

r

Express the magnitude of the force acting on the small sphere due to one cavity:

22

C

2⎟⎠⎞

⎜⎝⎛+

=Rd

GM'mF

where M′ is the negative mass of a cavity.

Relate the negative mass of a cavity to the mass of the sphere before hollowing: ( ) MR

RVM'

813

34

81

3

34

2

−=−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=−=

πρ

πρρ

Letting θ be the angle between the x axis and the line joining the center of the small sphere to the center of either cavity, use the law of gravity to express the force due to the two cavities:

iF ˆcos

48

22

2C θ

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=Rd

GMmr

because, by symmetry, the y components add to zero.

Express cosθ :

4

cos2

2 Rd

d

+

=θ

Gravity

889


i

iF

ˆ

44

ˆ

444

2/322

22

22

C

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

Rd

GMmd

Rd

dRd

GMmr


i

iiF

ˆ

4

41

ˆ

44

ˆ

2/322

3

2

2/322

2

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

+−=

Rd

d

dGMm

Rd

GMmdd

GMmr

(b) Evaluate F

rat d = R:

( )

i

iF

ˆ821.0

ˆ

4

41

2

2/322

3

2

RGMm

RR

R

RGMmR

−=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

−−=r

80 •• Picture the Problem Let R be the size of the cluster, and N the total number of stars in it. We can apply Newton’s law of gravity and the 2nd law of motion to relate the net force (which depends on the number of stars N(r) in a sphere whose radius is equal to the distance between the star of interest and the center of the cluster) acting on a star at a distance r from the center of the cluster to its speed. We can use the definition of density, in conjunction with the assumption of uniform distribution of the starts within the cluster, to find N(r) and, ultimately, express the orbital speed v of a star in terms of the total mass of the cluster. Using Newton’s law of gravity and 2nd law, express the force acting on a star at a distance r from the center of the cluster:

( )rvM

rMrGNrF

2

2

2

)( ==

where N(r) is the number of stars within a distance r of the center of the cluster and M is the mass of an individual star.

Chapter 11

890

Using the uniform distribution assumption and the definition of density, relate the number of stars N(r) within a distance r of the center of the cluster to the total number N of stars in the cluster:

( )3

343

34 R

NMrMrN

ππρ ==

or

( ) 3

3

RrNrN =


vMRr

rGNM 2

3

3

2

2

=

or

23

2

vRrGNM =


3RGNMrv = ⇒ rv ∝

center. thefrom distanceith linearly w increasescluster theofcenter thearoundorbit circular ain star a of ity mean veloc The

rv

General Problems

*81 • Picture the Problem We can use Kepler’s 3rd law to relate Pluto’s period to its mean distance from the sun.

Using Kepler’s 3rd law, relate the period of Pluto to its mean distance from the sun:

32 CrT =

where 3219

s

2

/ms102.9734 −×==GM

C π.

Solve for T: 3CrT =


( )

y249

d365.25y1

h24d1

s3600h1s01864.7

AUm101.50AU5.39/ms10973.2

9

3113219

=

××××=

⎟⎟⎠

⎞⎜⎜⎝

⎛ ×××= −T

Gravity

891

82 • Picture the Problem Consider an object of mass m at the surface of the earth. We can relate the weight of this object to the gravitational field of the earth and to the mass of the earth.

Using Newton’s 2nd law, relate the weight of an object at the surface of the earth to the gravitational force acting on it:

2E

E

RmGMmgw ==

Solve for ME: G

2E

EgRM =

Substitute numerical values and evaluate ME:

( )( )

kg1097.5

/kgmN106.6726m106.37kg/N9.81

24

2211

26

E

×=

⋅××

= −M

83 •• Picture the Problem The work you must do against gravity to move the particle from a distance r1 to r2 is the negative of the change in the particle’s gravitational potential energy. (a) Relate the work you must do to the change in the gravitational potential energy of the earth-particle system:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=−=∆−= ∫∫

21E

12E

2Eg

11

11

2

1

2

1

rrmGM

rrmGM

rdrmGMdrFUW

r

r

r

r

(b) Substitute 2

EgR for GME, RE for

r1, and RE + h for r2 to obtain: ⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=hRR

mgRWEE

2E

11 (1)

Chapter 11

892

(c) Rewrite equation (1) with a common denominator and simplify to obtain:

( )

mgh

Rhmgh

hRRmgh

hRRRhRmgRW

≈

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−+

=

E

E

E

EE

EE2E

1

1

when h << RE. 84 •• Picture the Problem The gravitational field outside a uniform sphere is given by

2rGMg −= and the field inside the sphere by ( ) .3 rRGMg −=

(a) Express g outside the sphere:

2rGMg −=

Find the mass of the sphere: ( )3

34 RVM πρρ ==

Substitute and simplify to obtain: ( )

2

3

34

2

334

rRG

rRGg ρπρ

−=−=


( ) ( ) ( )2

2

2

332211

34 /kgmN559.0m100kg/m2000kg/mN10673.6

rrg ⋅

−=⋅×

−=−

(b) Express the gravitational field inside the uniform sphere:

( )

Gr

rR

RGrR

GMg

ρπ

πρ

34

3

334

3

−=

−=−=


( )( ) ( )rrg mN/kg1059.5/kgmN106.6726kg/m2000 72211334 ⋅×−=⋅×−= −−π

85 •• Picture the Problem We can use Kepler’s 3rd law to relate the period of the satellite to its mean distance from the center of Jupiter.

Gravity

893

Use Kepler’s 3rd law to relate the period of the satellite to its mean distance from the center of Jupiter:

( )3JJ

22 4 hR

GMT +=

π

Solve for h: J

32

J2

4RGMTh −=

π (1)

Express the mass of Jupiter in terms of the mass of the earth:

EJ 320MM =

Express the volume of Jupiter in terms of the mass of the earth:

EJ 1320VV =

Express the volumes of Jupiter and Earth in terms of their radii and solve for RJ:

E3

J 1320RR =

Substitute in equation (1) to obtain: E

332

E2

13204320 RMGTh −=π

Express the period of the satellite in seconds:

s1054.3min

s60min50h

s3600h9

min50h9

4×=

×+×=

+=T


( ) ( ) ( )

( )m1096.8

m1037.613204

kg1098.5320/kgmN106726.6s1054.3

7

63

32

24221124

×=

×−

×⋅××=

−

πh

86 •• Picture the Problem Let m represent the mass of the spacecraft. From Kepler’s 3rd law we know that its period will be a minimum when it is in orbit just above the surface of the moon. We’ll use Newton’s 2nd law to relate the angular velocity of the spacecraft to the gravitational force acting on it.

Chapter 11

894

Relate the period of the spacecraft to its angular velocity:

ωπ2

=T (1)

Using Newton’s 2nd law of motion, relate the gravitational force acting on the spacecraft when it is in orbit at the surface of the moon to the angular velocity of the spacecraft:

∑ == 2M2

M

Mradial ωmR

RmGMF

Solve for ω: ( )

ρπ

ρπω

G

RRG

RGM

34

3M

3M3

4

3M

M

=

==

Substitute in equation (1) and simplify to obtain: GG

Tρπ

ρππ 32

34min ==

Substitute numerical values and evaluate Tmin:

( )( ) min 48h 1s6503kg/m3340/kgmN106726.6

332211min ==

⋅×= −

πT

87 •• Picture the Problem We can use conservation of energy to establish a relationship between the height h to which the projectile will rise and its initial speed. The application of Newton’s 2nd law will relate the orbital speed, which is equal to the initial speed of the projectile, to the mass and radius of the moon. Use conservation of energy to relate the initial energies of the projectile to its final energy:

0ifif =−+− UUKK

or, because Kf = 0,

0M

M

M

M221 =+

+−−

RmGM

hRmGMmv

Solve for h:

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−

= 1

21

1

M

M2

GMRv

Rh (1)

Use Newton’s 2nd law to relate velocity of the satellite to the ∑ ==

M

2

2M

Mradial R

vmR

mGMF

Gravity

895

gravitational force acting on it: Solve for v2:

M

M2

RGMv =

Substitute for v2 in equation (1) and simplify to obtain: Mm70.11

211

1==

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−

= RRh

*88 •• Picture the Problem If we assume the astronauts experience a constant acceleration in the barrel of the cannon, we can use a constant-acceleration equation to relate their exit speed (the escape speed from the earth) to the acceleration they would need to undergo in order to reach that speed. We can use conservation of energy to express their escape speed in terms of the mass and radius of the earth and then substitute in the constant-acceleration equation to find their acceleration. To find the balance point between the earth and the moon we can equate the gravitational forces exerted by the earth and the moon at that point. (a) Assuming constant acceleration down the cannon barrel, relate the ship’s speed as it exits the barrel to the length of the barrel and the acceleration required to get the ship to escape speed:

l∆= av 22e

where l is the length of the cannon.

Solve for the acceleration:

l∆=

2

2eva (1)

Use conservation of energy to relate the initial energy of astronaut’s ship to its energy when it has escaped the earth’s gravitational field:

0=∆+∆ UK or

0ifif =−+− UUKK

When the ship has escaped the earth’s gravitational field:

0ff ==UK and

0ii =−− UK or

0E2e2

1 =⎟⎠⎞

⎜⎝⎛−−−

RmGMmv

where m is the mass of the spaceship.

Solve for 2ev to obtain:

RGMv E2

e2

=

Chapter 11

896

Substitute in equation (1) to obtain: R

GMal∆

= E


( )( )

( )( )

g

a

300,23m/s1029.2

km6370m274kg105.98

/kgmN10673.6

25

24

2211

≈×=

××

⋅×= −

unlikely!extremely is Survival

(b) Let the distance from the center of the earth to the center of the moon be R, and the distance from the center of the spaceship to the earth be x. If M is the mass of the earth and m the mass of the moon, the forces will balance out when:

22 )( xRGm

xGM

−=

or

mxR

Mx −

=

where we’ve ignored the negative solution, as it doesn't indicate a point between the two bodies.


Mm

Rx+

=1

Substitute numerical values and evaluate x:

m1046.3

kg105.98kg107.361

m1084.3

8

24

22

8

×=

××

+

×=x

(c) anything. weigh toseemnot wouldso and

fall,-freein be wouldastronauts the trip,entire theDuring not. isit No

89 •• Picture the Problem Let the origin of our coordinate system be at the center of mass of the binary star system and let the distances of the stars from their center of mass be r1 and r2. The period of rotation is related to the angular velocity of the star system and we can use Newton’s 2nd law of motion to relate this velocity to the separation of the stars. Relate the square of the period of the motion of the stars to their angular velocity:

2

22 4

ωπ

=T (1)

Gravity

897

Using Newton’s 2nd law of motion, relate the gravitational force acting on the star whose mass is m2 to the angular velocity of the system:

( )∑ =+

= 2222

21

21radial ωrm

rrmGmF

Solve for ω2: ( )2

212

12

rrrGm+

=ω (2)

From the definition of the center of mass we have:

2211 rmrm = (3) where 21 rrr += (4)

Eliminate r1 from equations (3) and (4) and solve for r2:

21

12 mm

rmr+

=

Eliminate r2 from equations (3) and (4) and solve for r1:

21

21 mm

rmr+

=

Substitute for r1 and r2 in equation (2) to obtain:

( )3

212

rmmG +

=ω

Finally, substitute in equation (1) and simplify: ( ) ( )21

32

321

22 44

mmGr

rmmGT

+=

+=

ππ

90 •• Picture the Problem Because the two-particle system has zero initial energy and zero initial linear momentum; we can use energy and momentum conservation to obtain simultaneous equations in the variables r, v1 and v2. We’ll assume that initial separation distance of the particles and their final separation r is large compared to the size of the particles so that we can treat them as though they are point particles. Use conservation of energy to relate the speeds of the particles when their separation distance is r:

fi EE =

or

rmGmvmvm 212

22212

11210 −+= (1)

Use conservation of linear momentum to obtain a second relationship between the speeds of the particles and their masses:

fi pp =

or 22110 vmvm += (2)

Chapter 11

898

Solve equation (2) for v1 and substitute in equation (1) to obtain: r

mGmmmmv 21

1

22

222

2=⎟⎟

⎠

⎞⎜⎜⎝

⎛+ (3)

Solve equation (3) for v2:

( )21

21

22

mmrGmv+

=

Solve equation (2) for v1 and substitute for v2 to obtain: ( )21

22

12

mmrGmv+

=

*91 •• Picture the Problem We can find the orbital speeds of the planets from their distance from the center of mass of the system and the period of their motion. Application of Kepler’s 3rd law will allow us to express the period of their motion T in terms of the effective mass of the system … which we can find from its definition.

Express the orbital speeds of the planets in terms of their period T: T

Rv π2=

where R is the distance to the center of mass of the four-planet system.

Apply Kepler’s 3rd law to express the period of the planets: 3

eff

24 RGM

T π=

where Meff is the effective mass of the four planets.


RGM

RGM

Rv eff

3

eff

242

==ππ

The distance of each planet from the effective mass is: 2

aR =

Find Meff from its definition: MMMMM

11111

eff

+++=

and MM 4

1eff =

Substitute for R and Meff to obtain:

aGMv42

=

Gravity

899

92 •• Picture the Problem Let r represent the separation of the particle from the center of the earth and assume a uniform density for the earth. The work required to lift the particle from the center of the earth to its surface is the integral of the gravitational force function. This function can be found from the law of gravity and by relating the mass of the earth between the particle and the center of the earth to the earth’s mass. We can use the work-kinetic energy theorem to find the speed with which the particle, when released from the surface of the earth, will strike the center of the earth. Finally, the energy required for the particle to escape the earth from the center of the earth is the sum of the energy required to get it to the surface of the earth and the kinetic energy it must have to escape from the surface of the earth. (a) Express the work required to lift the particle from the center of the earth to the earth’s surface:

∫=E

0

R

FdrW (1)

where F is the gravitational force acting on the particle.

Using the law of gravity, express the force acting on the particle as a function of its distance from the center of the earth:

2rGmMF = (2)

where M is the mass of a sphere whose radius is r.

Express the ratio of M to ME: ( )( )3

E34

334

E Rr

MM

πρπρ

= ⇒ 3E

3

E RrMM =

Substitute for M in equation (2) to obtain:

rRmgr

RmgRr

RGmMF

E3E

2E

3E

E ===

Substitute for F in equation (1) and evaluate the integral: 2

E

0E

E gmRrdrRmgW

R

== ∫

(b) Use the work-kinetic energy theorem to relate the kinetic energy of the particle as it reaches the center of the earth to the work done on it in moving it to the surface of the earth:

221 mvKW =∆=

Substitute for W and solve for v: EgRv =

Chapter 11

900

(c) Express the total energy required for the particle to escape when projected from the center of the earth:

2esc2

1

2e2

1esc

mv

mvWE

=

+=

where ve is the escape speed from the surface of the earth.

Substitute for W and solve for vesc: Eesc 3gRv =

Substitute numerical values and evaluate vesc:

( )( )km/s7.13

m106.37N/kg9.813 6esc

=

×=v

93 •• Picture the Problem We need to find the gravitational field in three regions: r < R1, R1 < r < R2, and r > R2. For r < R1: 0=g

For r > R2, g(r) is the field of a mass M centered at the origin:

( ) 2rGMrg =

For R1 < r < R2, g(r) is determined by the mass within the shell of radius r:

( ) 2rGmrg = (1)

where ( )31

334 Rrm −= πρ (2)

Express the density of the spherical shell: ( )3

1323

4 RRM

VM

−==

πρ

Substitute for ρ in equation (2) and simplify to obtain:

( )31

32

31

3

RRRrMm

−−

=

Substitute for m in equation (1) to obtain: ( ) ( )

( )31

32

2

31

3

RRrRrGM

rg−−

=

Gravity

901

A graph of gr with R1 = 2, R2 = 3, and GM = 1 follows:

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0 2 4 6 8

r

g r

94 •• Picture the Problem A ring of radius R is shown to the right. Choose a coordinate system in which the origin is at the center of the ring and x axis is as shown. An element of length dL and mass dm is responsible for the field dg at a distance x from the center of the ring. We can express the x component of dg and then integrate over the circumference of the ring to find the total field as a function of x. (a) Express the differential gravitational field at a distance x from the center of the ring in terms of the mass of elemental length dL:

22 xRGdmdg+

=

Relate the mass of the element to its length:

dLdm λ=

where λ is the linear density of the ring.

Substitute to obtain: 22 xR

dLGdg+

=λ

By symmetry, the y and z components of g vanish. Express the x component of dg:

θλθ

cos

cos

22 xRdLG

dgdg x

+=

=

Chapter 11

902

Referring to the figure, express cosθ : 22

cosxR

x+

=θ


( ) 2/3222222 xRxdLG

xRx

xRdLGdgx

+=

+×

+=

λλ

Because R

Mπ

λ2

= : ( ) 2/3222 xRR

xdLGMdg x+

=π

Integrate to find g(x): ( )

( )

( )x

xR

GM

dLxRR

xGMxgR

2/322

2

02/3222

+=

+= ∫

π

π

A plot of gx is shown below. The curve is normalized for R = 1 and GM = 1.

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0 1 2 3 4

x

g x

(b) Differentiate g(x) with respect to x and set the derivative equal to zero to identify extreme values:

( ) ( )( ) ( ) ( ) extremafor 022/122

322232/322

=⎥⎥⎦

⎤

⎢⎢⎣

⎡+

+

−+= xRx

xRxRxGM

dxdg

Simplify to obtain: ( ) ( ) 03 2/12222/322 =+−+ RxxRx

Gravity

903


2Rx ±=

Because the curve is concave downward, we can conclude that this result corresponds to a maximum. Note that this result agrees with our graphical maximum.

95 ••• Picture the Problem The diagram shows a segment of the wire of length dx and mass dm = λdx at a distance x from the origin of our coordinate system. We can find the magnitude of the gravitational field at a distance r from the wire from the resultant gravitational force acting on a particle of mass m′ located at point P and then integrating over the length of the wire. Express the gravitational force acting on a particle of mass m′ at a distance r from the wire due to the segment of the wire of length dx:

m'dgdF = or

m'dFdg =

Using Newton’s law of gravity, express dF:

2RdxGm'dF λ

=

or, because 222 rxR += ,

22 rxdxGm'dF

+=

λ

Substitute and simplify to express the gravitational field due to the segment of the wire of length dx:

22 rxdxGdg+

=λ

By symmetry, the segment on the opposite side of the origin at the same distance from the origin will cancel out all but the radial component of the field, so the gravitational field will be given by:

( ) dxrxrG

rxr

rxdxGrx

dxGdg

2322

2222

22 cos

+=

++=

+=

λ

λ

θλ

Integrate dg from x′ = −∞ to x′ = +∞ to obtain:

( ) ( ) rG

rxx

rGdx

rxrGdx

rxrGg λλλλ 2

'2'

'2'

' 022

02/3222/322

=⎥⎦

⎤⎢⎣

⎡

+=

+=

+=

∞∞∞

∞−∫∫

Chapter 11

904

96 ••• Picture the Problem We can use the relationship between the angular velocity of an orbiting object and its tangential velocity to express the speeds vin and vout of the innermost and outermost portions of the ring. In part (b) we can use Newton’s law of gravity, in conjunction with the 2nd law of motion, to relate the tangential speed of a chunk of the ring to the gravitational force acting on it. As in part (a), once we know vin and vout, we can express the difference between them to obtain the desired results. (a) Express the speed of a point in the ring at a distance R′ from the center of the planet under the assumption that the ring is solid and rotates with an angular velocity ω:

( ) RRv ω='

Express the speeds vin and vout of the innermost and outermost portions of the ring:

( )ωrRv 21

in −= and

( )ωrRv 21

out +=

Express the difference between vout and vin:

( ) ( )

Rrvr

Rvr

rRrRvv

===

−−+=−

ω

ωω 21

21

inout

(b) Assume that a chunk of the ring is moving in a circular orbit around the center of the planet under the force of gravity. Then, we can find its velocity by equating the force of gravity to the centripetal force needed to keep it in orbit:

''

2

2 Rmv

RGMm

=

or

'RGMv =

where M is the mass of the planet and R' the distance from the center.

Substitute for R′ to express vout:

21

21out

211

211

−

⎟⎠⎞

⎜⎝⎛ +=

⎟⎠⎞

⎜⎝⎛ +

=+

=

Rr

RGM

RrR

GMrR

GMv

Expand binomially to obtain:

)

⎟⎠⎞

⎜⎝⎛ −≈

+

⎜⎝⎛ −=

Rr

RGM

Rr

RGMv

411

sorder termhigher 21

211out

Gravity

905

Proceed similarly to obtain, for vin: ⎟⎠⎞

⎜⎝⎛ +≈

Rr

RGMv

411in

Express the difference between vout and vin:

⎟⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −≈−

Rr

RGM

Rr

RGM

Rr

RGMvv

21

411

411inout

and, because R

GMv = , vRrvv

21

inout −≈−

97 ••• Picture the Problem Let U = 0 at x = ∞. The potential energy of an element of the stick dm and the point mass m0 is given by the definition of gravitational potential energy:

rdmGmdU 0−= where r is the separation of dm and m0.

(a) Express the potential energy of the masses m0 and dm: xx

dmGmdU

−−=

0

0

The mass dm is proportional to the size of the element dx:

dxdm λ=

where LM

=λ .

Substitute these results to express dU in terms of x: ( )xxL

dxGMmxxdxGmdU

−−=

−−=

0

0

0

0λ

(b) Integrate to find the total potential energy for the system:

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ −=

−−= ∫

−

22ln

2ln

2ln

0

00

000

2/

2/ 0

0

LxLx

LGMm

LxLxL

GMmxx

dxL

GMmUL

L

(c) Because x0 is a general point along the x axis:

( )⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

+=−=

2

1

2

1

00

0

00 LxLxL

GmmdxdUxF

Chapter 11

906

Simplify this expression to obtain: ( )422

00 Lx

GmmxF−

−=

in agreement with the result of Example 11-8.

*98 ••• Picture the Problem Choose a mass element dm of the rod of thickness dx at a distance x from the origin. All such elements of the rod experience a gravitational force dF due to presence of the sphere centered at the origin. We can find the total gravitational force of attraction experienced by the rod by integrating dF from x = a to x = a + L. Express the gravitational force dF acting on the element of the rod of mass dm:

2xGMdmdF =

Express dm in terms of the mass m and length L of the rod:

dxLmdm =

Substitute to obtain: 2x

dxL

GMmdF =

Integrate dF from x = a to x = a + L to find the total gravitational force acting on the rod:

( )LaaGMm

xLGMmdxx

LGMmF

La

a

La

a

+=

⎥⎦⎤

⎢⎣⎡−==

++−∫

12

99 ••• Picture the Problem The semicircular rod is shown in the figure. We’ll use an

element of length θπ

θ dLRd = whose

mass dM is θπ

dM. By symmetry, 0=yF .

We’ll first find dFx and then integrate over θ from −π/2 to π/2. Express dFx: θθ

ππ

cos22 dL

GMmR

GmdMdFx

⎟⎠⎞

⎜⎝⎛

==

Gravity

907

Integrate dFx over θ from −π/2 to π/2: 2

2/

2/2

2cosLGMmd

LGMmFx

πθθπ π

π

== ∫−

Substitute numerical values and evaluate Fx:

( )( )( )( )

pN5.33m5

kg0.1kg20/kgmN106.672622

2211

=⋅×

=−πFx

*100 ••• Picture the Problem We can begin by expressing the forces exerted by the sun and the moon on a body of water of mass m and taking the ratio of these forces. In (b) we’ll simply follow the given directions and in (c) we can approximate differential quantities with finite quantities to establish the given ratio. (a) Express the force exerted by the sun on a body of water of mass m: 2

S

SS r

mGMF =

Express the force exerted by the moon on a body of water of mass m: 2

m

mm r

mGMF =

Divide the first of these equations by the second and simplify to obtain:

2Sm

2mS

m

S

rMrM

FF

=

Substitute numerical values and evaluate this ratio:

( )( )( )( )177

m101.50kg107.36m103.84kg101.99

21122

2830

m

S

=

××

××=

FF

(b) Find drdF

: rF

rmGm

drdF 22

321 −=−=

Solve for the ratio F

dF:

rdr

FdF 2−=

(c) Express the change in force ∆F for a small change in distance ∆r:

rrFF ∆−=∆ 2

Chapter 11

908

Express SF∆ :

S3S

S

SS

2S

S

S

2

2

rr

GmM

rrr

GmM

F

∆−=

∆−=∆

Express mF∆ : m3

m

mm 2 r

rGmM

F ∆−=∆

Divide the first of these equations by the second and simplify:

3Sm

3mS

m

S3

Sm

3mS

m3m

m

S3S

S

m

S

rMrM

rr

rMrM

rr

M

rrM

FF

=

∆∆

=∆

∆=

∆∆

because .1m

S =∆∆

rr

Substitute numerical values and evaluate this ratio:

( )( )( )( )

454.0

m101.50kg107.36m103.84kg101.99

31122

3830

m

S

=

××

××=

∆∆

FF

101 •• Picture the Problem Let MNS be the mass of the Neutron Star and m the mass of each robot. We can use Newton’s law of gravity to express the difference in the tidal-like forces acting on the coupled robots. Expanding the expression for the force on the robot further from the Neutron Star binomially will lead us to an expression for the distance at which the breaking tension in the connecting cord will be exceeded.

(a) stressed. is cable theseparation thisopposingIn separate. they would

and robot,upper theofan that greater th be on wouldaccelerati its cable thefornot it were if so robot,lower on thegreater is force nalgravitatio The

Gravity

909

(b) Letting the separation of the two robots be ∆r, and the distance from the center of the star to the lower robot be r, use Newton’s law of gravity to express the difference in the forces acting on the robots:

( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ∆+−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ ∆+

−=

∆+−=

−2

2NS

22

2NS

2N

2NS

tide

11

1

11

rr

rmGM

rrr

rmGM

rrmGM

rmGMF S

Expand the expression in the square brackets binomially to obtain:

rr

rr

rr

∆=

⎟⎠⎞

⎜⎝⎛ ∆−−≈⎟

⎠⎞

⎜⎝⎛ ∆+−

−

2

211112


rr

mGMF ∆≈ 3NS

tide2

Letting FB be the breaking tension of the cord, substitute for Ftide and solve for the value of r corresponding to the breaking strain being exceeded:

3 NS2 rF

mGMrB

∆=

Substitute numerical values and evaluate r:

( )( )( ) ( ) km220m1kN25

kg1kg1099.1/kgmN10673.623

302211

=×⋅×

=−

r

Chapter 11

910

911

Chapter 12 Static Equilibrium and Elasticity Conceptual Problems 1 • (a) False. The conditions ∑ =

i i 0Fr

and ∑ =i i 0τr must be satisfied.

(b) True. The necessary and sufficient conditions for static equilibrium are ∑ =

i i 0Fr

and

∑ =i i 0τr .

(c) True. The conditions ∑ =

i i 0Fr

and ∑ =i i 0τr must be satisfied.

(d) False. An object is in equilibrium provided the conditions ∑ =

i i 0Fr

and ∑ =i i 0τr are

satisfied.

2 • False. The location of the center of gravity depends on the mass distribution. 3 • No. The definition of the center of gravity does not require that there be any material at its location. 4 • Determine the Concept When the acceleration of gravity is not constant over an object, the center of gravity is the pivot point for balance. 5 •• Determine the Concept This technique works because the center of mass must be directly under the balance point. Thus, a line drawn straight downward will pass through the center of mass, and another line drawn straight downward when the figure is hanging from another point will also pass through the center of mass. The center of mass is where the lines cross.

*6 • Determine the Concept No. Because the floor can exert no horizontal force, neither can the wall. Consequently, the friction force between the wall and the ladder is zero regardless of the coefficient of friction between the wall and the ladder.

Chapter 12

912

7 • Determine the Concept We know that equal lengths of aluminum and steel wire of the same diameter will stretch different amounts when subjected to the same tension. Also, because we are neglecting the mass of the wires, the tension in them is independent of which one is closer to the roof and depends only on W. correct. is )(b

8 • Determine the Concept Yes; if it were otherwise, angular momentum conservation would depend on the choice of coordinates. *9 • Determine the Concept The condition that the bar is in rotational equilibrium is that the net torque acting on it be zero; i.e., R1M1 = R2M2. This condition is satisfied provided R1 = R2 and M1 = M2. correct. is )(c

10 •• Determine the Concept You cannot stand up because your body’s center of gravity must be above your feet. *11 •• Determine the Concept The tensile strengths of stone and concrete are at least an order of magnitude lower than their compressive strengths, so you want to build compressive structures to match their properties. Estimation and Approximation 12 •• Picture the Problem The diagram to the right shows the forces acting on the crate as it is being lifted at its left end. Note that when the crowbar lifts the crate, only half the weight of the crate is supported by the bar. Choose the coordinate system shown and let the subscript ″pb″ refer to the pry bar. The diagram below shows the forces acting on the pry bar as it is being used to lift the end of the crate.

Static Equilibrium and Elasticity

913

Assume that the maximum force F ′ you can apply is 500 N (about 110 lb). Let l be the distance between the points of contact of the steel bar with the floor and the crate, and let L be the total length of the bar. Lacking information regarding the bend in pry bar at the fulcrum, we’ll assume that it is small enough to be negligible. We can apply the condition for rotational equilibrium to the pry bar and a condition for translational equilibrium to the crate when its left end is on the verge of lifting. Apply ∑ = 0yF to the crate:

0npb =+− FWF (1)

Apply 0=∑τr to the crate about

an axis through point B and perpendicular to the plane of the page to obtain:

021

n =− wWwF

Solve for Fn: WF 21

n =

as noted in Picture the Problem.

Solve equation (1) for Fpb and substitute for Fn to obtain:

WWWF 21

21

pb =−=

Apply 0=∑τr to the pry bar about

an axis through point A and perpendicular to the plane of the page to obtain:

( ) 0pb =−− FLF ll

Solve for L: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

FF

L pb1l

Substitute for Fpb to obtain:

⎟⎠⎞

⎜⎝⎛ +=

FWL2

1l

Chapter 12

914


( ) ( ) cm0.55N5002N45001m1.0 =⎟⎟

⎠

⎞⎜⎜⎝

⎛+=L

*13 •• Picture the Problem We can derive this expression by imagining that we pull on an area A of the given material, expressing the force each spring will experience, finding the fractional change in length of the springs, and substituting in the definition of Young’s modulus. (a) Express Young’s modulus: LL

AFY∆

= (1)

Express the elongation ∆L of each spring: k

FL s=∆ (2)

Express the force Fs each spring will experience as a result of a force F acting on the area A:

NFF =s

Express the number of springs N in the area A: 2a

AN =


AFaF

2

s =

Substitute in equation (2) to obtain, for the extension of one spring: kA

FaL2

=∆

Assuming that the springs extend/compress linearly, express the fractional extension of the springs:

kAFa

kAFa

aaL

LL

==∆

=∆ 2

tot 1


ak

kAFaAF

Y ==

(b) From our result in part (a): Yak =

From Table 12-1: 2112 N/m102GN/m200 ×==Y

Assuming that a ~ 1 nm, evaluate k: ( )( ) N/m200m10N/m102 9211 =×= −k


915

Conditions for Equilibrium 14 • Picture the Problem Let w1 represent the weight of the 28-kg child sitting at the left end of the board, w2 the weight of the 40-kg child, and d the distance of the 40-kg child from the pivot point. We can apply the condition for rotational equilibrium to find d.

Apply 0=∑τr about an axis through

the pivot point P:

( ) 02 21 =− dwmw

Solve for and evaluate d: ( ) ( ) ( )( ) m4.1

kg40m2kg28m2

2

1 ===g

gw

wd

15 • Picture the Problem Let F1 represent the force exerted by the floor on Misako’s feet, F2 the force exerted on her hands, and m her mass. We can apply the condition for rotational equilibrium to find F2.

Apply 0=∑τr about an axis

through point 0:

( ) ( ) 0m9.0m5.12 =−mgF

Solve for F2: ( )m5.1

m9.02

mgF =

Substitute numerical values and evaluate F2:

( )( )( )

N318

m5.1m9.0m/s81.9kg54 2

2

=

=F

*16 • Picture the Problem Let F represent the force exerted by Misako’s biceps. To find F we apply the condition for rotational equilibrium about a pivot chosen at the tip of her elbow.

Apply 0=∑τr about an axis ( ) ( )( ) 0N18cm28cm5 =−F

Chapter 12

916

through the pivot: Solve for F: ( )( ) N101

cm5N18cm28

==F

17 • Picture the Problem Choose a coordinate system in which upward is the positive y direction and to the right is the positive x direction and use the conditions for translational equilibrium.

(a) Apply 0=∑F

r to the forces

acting on the tip of the crutch:

0sincs =+−=∑ θFfFx (1)

and

∑ =−= 0coscn θFFFy (2)

Solve equation (2) for Fn and assuming that fs = fs,max, obtain:

θµµ coscsnsmaxs,s FFff ===

Substitute in equation (1) and solve for µs:

θµ tans =

(b) strides. longfor large is

becausefriction static oft coefficien large a requires strides long Takingθ

(c) slipping. avoid tosmall bemust surface, on the ice is therei.e., small, is If s θµ

The Center of Gravity 18 • Picture the Problem Let the weight of the automobile be w. Choose a coordinate system in which the origin is at the point of contact of the front wheels with the ground and the positive x axis includes the point of contact of the rear wheels with the ground. Apply the definition of the center of gravity to find its location.

Use the definition of the center of gravity:

( ) ( )( )w

ww

xwWxi

ii

m84.0m242.0058.0

cg

=+=

= ∑

or, because W = w, ( ) ( )wwx m84.0cg =


917

Solve for xcg: m84.0cg =x

*19 • Picture the Problem The figures are shown on the right. The center of mass for each is indicated by a small +. At static equilibrium, the center of gravity is directly below the point of support.

20 •• Picture the Problem Using the coordinate system indicated in the figure, we can apply the definition of the center of gravity to determine xcg and ycg. Apply the definition of the center of gravity to find xcg: ( )( ) ( )( )

( )( ) ( )( )( )a

aaaa

xwWxi

ii

N170N50N30

N60N40

23

23

21

21

cg

=++

+=

= ∑

or, because W = 180 N, ( ) ( )ax N170N180cg =

Solve for xcg: aax 944.0

N180N170

cg ==

Apply the definition of the center of gravity to find ycg: ( )( ) ( )( )

( )( ) ( )( )( )a

aaaa

ywWyi

ii

N180N50N30

N60N40

21

23

23

21

cg

=++

+=

= ∑

or, because W = 180 N, ( ) ( )ay N180N180cg =

Solve for ycg: ay =cg

The coordinates of the center of gravity are:

( ) ( )aayx ,944.0, cgcg =

Chapter 12

918

21 •• Picture the Problem Let the origin of the coordinate system be at the lower left corner of the plate and the positive x direction be to the right. Let a and b be the length and width of the plate. Let σ be the mass per unit area of the plate. Then the weight of the plate is given by w = abσg and that of the matter missing from the hole is .2 gR σπ− Noting that, by

symmetry, ycg = b/2, we can apply the definition of the center of gravity to find xcg.

Apply the definition of the center of gravity to find xcg:

( )( ) ( )( )RagRagab

xwWxi

ii

−−=

= ∑σπσ 2

21

cg

or, because ,2

holeplate gRgabwwW σπσ −=−=

( ) ( )( )( )( )RaR

agabgRgabx

−−

=−

σπ

σσπσ2

212

cg

Solve for xcg: 2

32221

cg RabRaRbax

πππ

−+−

=

The coordinates of the center of gravity are: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛−

+−= b

RabRaRbayx 2

12

32221

cgcg ,,π

ππ

Some Examples of Static Equilibrium 22 • Picture the Problem We can use the given definition of the mechanical advantage of a lever and the condition for rotational equilibrium to show that M = x/X.

(a) Express the definition of mechanical advantage for a lever: f

FM =

Apply the condition for rotational equilibrium to the lever:

0=− XFxf

Solve for the ratio of F to f to obtain: X

xfF=

Substitute to obtain: X

xM =

(b) force.

applied theofmovement short a using distance large aover load themove to wishesone when useful is force applied for the armmoment shorter A


919

23 • Picture the Problem The force diagram shows the tension in the forestay, ,FT

r the

tension in the backstay, ,BTr

the

gravitational force on the mast ,grm and the force exerted

by the deck, .DFr

Let the origin of the

coordinate system be at the foot of the mast with the positive x direction to the right and the positive y direction upward. Because the mast is in equilibrium, we can apply the conditions for both translational and rotational equilibrium to find the tension in the backstay and the force that the deck exerts on the mast.

Apply 0=∑τr to the mast about an axis

through its foot and solve for TB:

( )( )( ) 045sinm88.4

sinN1000m88.4

B

F

=°− Tθ

and ( )

°=

45sinsinN1000 F

BθT

Find θF, the angle of the forestay with the vertical: °=⎟⎟

⎠

⎞⎜⎜⎝

⎛= − 3.29

m4.88m2.74tan 1

Fθ

Substitute to obtain: ( ) N692

45sin3.29sinN1000

B =°

°=T

Apply the condition for translational equilibrium in the x direction to the mast: 0sin

45sincos

FF

BD

=−

°+=∑θ

θ

TTFFx

or ( )( )

0sin45N692

sin29.3N1000cosD

≈°−°=θF

Apply the condition for translational equilibrium in the y direction to the mast: 045cos

cossin

B

FFD

=−°−

−=∑mgT

TFFy θθ

or

Chapter 12

920

( )( )( )( )N2539

m/s9.81kg120cos45N692

cos29.3N1000sin

2

D

=+

°+°=θF

Because FDcosθ = 0: °= 90θ , kN54.2D =F

and

moving. frommast eprevent th torequired isblock no

24 •• Picture the Problem The diagram shows

,gr

M the weight of the beam, ,gr

m the

weight of the student, and the force the ledge exerts ,F

r acting on the beam.

Because the beam is in equilibrium, we can apply the condition for rotational equilibrium to the beam to find the location of the pivot point P that will allow the student to walk to the end of the beam.


through the pivot point P:

( ) 0m5 =−− mgxxMg

Solve for x: ( ) m4.17kg60kg300

kg30055=

+=

+=

mMMx

*25 •• Picture the Problem The diagram shows

,wr

the weight of the student, ,PFr

the

force exerted by the board at the pivot, and ,sF

r the force exerted by the scale, acting

on the student. Because the student is in equilibrium, we can apply the condition for rotational equilibrium to the student to find the location of his center of gravity.

Apply 0=∑τr about an axis ( ) 0m2s =−wxF


921

through the pivot point P: Solve for x: ( )

wFx sm2

=


( )( )( )( ) m728.0

m/s9.81kg70N250m2

2 ==x

26 •• Picture the Problem The diagram shows

,gr

m the weight of the board, ,HFr

the force exerted by the hinge, ,g

rM the weight of

the block, and ,Fr

the force acting

vertically at the right end of the board. Because the board is in equilibrium, we can apply the condition for rotational equilibrium to it to find the magnitude of

.Fr

(a) Apply 0=∑τr about an axis

through the hinge:

( )[ ] ( )[ ]( )[ ] 030cosm8.0

30cosm5.130cosm3=°−°−°

MgmgF

Solve for F: ( ) ( ) gMmF

m3m8.0m5.1 +

=


( )( ) ( )( )

( )N181

m/s81.9m3

m0.8kg60m1.5kg5

2

=

×

+=F

(b) Apply ∑ = 0yF to the board:

0H =+−− FmgMgF

Solve for and evaluate FH: ( )( )( )

N457

N181m/s9.81kg5kg60 2H

=

−+=

−+=−+= FgmMFmgMgF

Chapter 12

922

(c) The force diagram showing the force F

r acting at right angles to the

board is shown to the right:

Apply 0=∑τr about the hinge: ( ) ( )[ ]( )[ ] 030cosm8.0

30cosm5.1m3=°−

°−Mg

mgF

Solve for F: ( ) ( )

°+

= 30cosm3

m8.0m5.1 gMmF


( )( ) ( )( )

( )N157

30cosm/s81.9m3

m0.8kg60m1.5kg5

2

=

°×

+=F

Apply ∑ = 0yF to the board: 030cossinH =°+−− FmgMgF θ

or ( ) °−+= 30cossinH FgmMF θ (1)

Apply ∑ = 0xF to the board: 030sincosH =°− FF θ

or °= 30sincosH FF θ (2)


( )°

°−+=

30sin30cos

cossin

H

H

FFgmM

FF

θθ

Solve for θ: ( )

⎥⎦⎤

⎢⎣⎡

°°−+

= −

30sin30costan 1

FFgmMθ

Substitute numerical values and evaluate θ :


923

( )( ) ( )( ) °=⎥

⎦

⎤⎢⎣

⎡°

°−= − 1.81

sin30N157cos30N157m/s9.81kg65tan

21θ

Substitute numerical values in equation (2) and evaluate FH:

( ) N5071.81cos

30sinN157H =

°°

=F

*27 • Picture the Problem The planes are frictionless; therefore, the force exerted by each plane must be perpendicular to that plane. Let 1F

rbe the force exerted by the 30°

plane, and let 2Fr

be the force exerted by the

60° plane. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Because the cylinder is in equilibrium, we can use the conditions for translational equilibrium to find the magnitudes of 1F

rand

2Fr

.

Apply ∑ = 0xF to the cylinder:

060sin30sin 21 =°−° FF (1)

Apply ∑ = 0yF to the cylinder:

060cos30cos 21 =−°+° WFF (2)

Solve equation (1) for F1:

21 3FF = (3)


060cos30cos3 22 =−°+° WFF

Solve for F2: ( ) WF =°+° 260cos30cos3

or

WWF 21

2 60cos30cos3=

°+°=

Substitute in equation (3): ( ) WWF 2

321

1 3 ==

Chapter 12

924

28 •• Picture the Problem The force diagram shows the forces ,HF

r ,2Tr

and 1Tr

acting

on the strut. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Because the strut is in equilibrium, we can apply the conditions for translational and rotational equilibrium to it.

(a) hinge. by thestrut on the exerted force

the, and and tensions thearestrut on the acting forces The H21 FTTrrr

(b) Apply 0=∑τr about an axis

through the hinge:

030sin 12 =−° ll TT

Solve for T1: 122v 30sin TTT =°=

or, because T1 = 80 N, N802v =T

(c) Apply ∑ = 0xF to the beam: 030coscos 2H =°−TF θ

or °= 30coscos 2H TF θ (1)

Apply ∑ = 0yF to the beam: 030sinsin 12H =−°+ TTF θ

or

°−=°−=30sinN80

30sinsin

2

21H

TTTF θ

(2)

Divide equation (2) by equation (1) to obtain: °

°−=

30cos30sinN80tan

2

2

TTθ

Solve for θ :

⎥⎦

⎤⎢⎣

⎡°

°−= −

30cos30sinN80tan

2

21

TTθ

Express T2 in terms of T2v: N160

30sinN80

30sin2v

2 =°

=°

=TT


925

Evaluate θ: ( )( ) °=⎥

⎦

⎤⎢⎣

⎡°

°−= − 0

30cosN16030sinN160N80tan 1θ

Substitute numerical values in equation (1) and evaluate FH:

( ) N1390cos

30cosN160H =

°°

=F to the

right.

29 •• Picture the Problem The force diagram shows the weight of the pirate, ,Mg

r the

weight of the victim, ,mgr

and the force

the deck exerts at the edge of the ship, F

racting at the fulcrum P. The

diagram also shows, for part (b), the weight of the plank acting through the plank’s center of gravity.

(a) Apply 0=∑τr at the pivot point P: ( ) 0m8 =−− mgxxMg

or ( ) 0m8 =−− mxxM

Solve for x: ( ) m5.00

kg63kg105kg10588

=+

=+

=mM

Mx


through the pivot point P:

( ) ( ) 0m4-m8 p =−−+ mgxxgmxMg

or ( ) ( ) 0m4m8 p =−−+− mxxmxM

Solve for x:

p

p48mmM

mMx

+++

=


( ) ( ) m87.4kg25kg63kg105

kg254kg1058=

+++

=x

Chapter 12

926

30 •• Picture the Problem The drawing shows the door and its two supports. The center of gravity of the door is 0.8 m above (and below) the hinge, and 0.4 m from the hinges horizontally. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Denote the horizontal and vertical components of the hinge force by FHh and FHv. Because the door is in equilibrium, we can use the conditions for translational and rotational equilibrium to determine the horizontal forces exerted by the hinges.


the lower hinge:

( ) ( ) 0m4.0m6.1Hh =−mgF

Solve for FHh: ( )m6.1

m4.0Hh

mgF =

Substitute numerical values and evaluate FHh:

( )( )( )

N44.1

m1.6m0.4m/s9.81kg18 2

Hh

=

=F

Apply ∑ = 0xF to the door and

solve for Hh'F :

0' HhHh =−FF

and N1.44'Hh =F

Note that the upper hinge pulls on the door

and the lower hinge pushes on it.


927

31 •• Picture the Problem The figure shows the wheel on the verge of rolling over the edge of the step. Note that, under this condition, the normal force the floor exerts on the wheel is zero. Choose the coordinate system shown in the figure and apply the conditions for translational equilibrium and the result for F from Example 12-4 to the wheel.

Apply 0=∑Fr

to the wheel: 01x =−=∑ FFFx

and

∑ =−= 01y MgFFy

Write 1F

rin vector form:

ji

jiFˆˆ

ˆˆ1y1x1

MgF

FF

+−=

+−=r

From Example 12-4 we have: ( )hR

hRhMgF−

−=

2

Substitute to obtain: ( )

( )ji

jiF

ˆˆ2

ˆˆ21

MgRh

hRhMg

MghR

hRhMg

+−

−=

+−

−−=

r

32 •• Picture the Problem The diagram shows the forces 1F

rand 2F

racting at the supports,

the weight of the board ,grm acting at its

center of gravity, and the weight of the diver g

rM acting at the end of the diving

board. Because the board is in equilibrium, we can apply the condition for rotational equilibrium to find the forces at the supports.


the left support:

( ) ( ) ( ) 0m2.4m1.2m2.1 2 =−− MgmgF

Chapter 12

928

Solve for F2: ( ) ( )( ) gMmF

m2.1m2.4m1.2

2+

=


( )( ) ( )( ) ( ) ncompressio kN,92.2m/s9.81

m2.1kg) (70m2.4kg30m1.2 2

2 =+

=F


the right support:

( ) ( ) ( ) 0m3m9.0m2.1 1 =−− MgmgF

Solve for F1: ( ) ( )( ) gMmF

m2.1m3m9.0

1+

=


( )( ) ( )( ) ( ) tensionkN,94.1m/s9.81

m2.1kg) (70m3kg30m9.0 2

1 =+

=F

33 •• Picture the Problem Let T be the tension in the line attached to the wall and L be the length of the strut. The figure includes w, the weight of the strut, for part (b). Because the strut is in equilibrium, we can use the conditions for both rotational and translational equilibrium to find the force exerted on the strut by the hinge.

(a) Express the force exerted on the strut at the hinge:

jiF ˆˆvh FF +=

r (1)

Ignoring the weight of the strut, apply 0=∑τr at the hinge:

( ) 045cos =°− WLLT

Solve for the tension in the line: ( )N43.42

45cosN6045cos=

°=°=WT

Apply 0=∑F

r to the strut: ∑ =°−= 045cosh TFFx


929

and

∑ =−°+= 045cosv MgTFFy

Solve for Fh: ( )

N0.3045cosN43.4245cosh

=°=°= TT

Solve for Fv:

( )N30.0

cos45N42.43N0645cosv

=°−=

°−= TMgF

Substitute in equation (1) to obtain: ( ) ( ) jiF ˆN0.30ˆN0.30 +=

r

(b) Including the weight of the strut, apply 0=∑τr at the hinge:

( ) 045cos2

45cos =⎟⎠⎞

⎜⎝⎛−°− wLWLLT

Solve for the tension in the line: ( )

( )( ) ( )

N5.49

N2045cos21N6045cos

45cos2145cos

=

⎟⎠⎞

⎜⎝⎛ °+°=

⎟⎠⎞

⎜⎝⎛ °+°= wWT

Apply 0=∑F

r to the strut: ∑ =°−= 045cosh TFFx

and

∑ =−−°+= 045cosv wWTFFy

Solve for Fh: ( )

N0.3545cosN5.4945cosh

=°=°= TT

Solve for Fv:

( )N.054

cos45N5.94N20N0645cosv

=°−+=

°−+= TwWF

Substitute in equation (1) to obtain: ( ) ( ) jiF ˆN0.45ˆN0.35 +=

r

Chapter 12

930

34 •• Picture the Problem Note that if the 60-kg mass is at the far left end of the plank, T1 and T2 are less than 1 kN. Let x be the distance of the 60-kg mass from T1. Because the plank is in equilibrium, we can apply the condition for rotational equilibrium to relate the distance x to the other distances and forces.


through the left end of the plank:

( ) ( ) ( )0

m5.2m4m5

J

pb2

=−

−−

gxm

gmgmT

Solve for x: ( ) ( ) ( )

gmgmgmT

xJ

pb2 m2m4m5 −−=


( )kN5886.0

mkN63.3m5 2 ⋅−=

Tx

Set T2 = 1 kN and evaluate x: ( )( ) m33.2kN5886.0

mkN63.3kN1m5=

⋅−=x

and m.33.20for safe is Julie << x


931

35 •• Picture the Problem The figure to the right shows the forces acting on the cylinder. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Because the cylinder is in equilibrium, we can apply the conditions for translational and rotational equilibrium to find Fn and the horizontal and vertical components of the force the corner of the step exerts on the cylinder.

(a) Apply 0=∑τr to the cylinder

about the step’s corner:

( ) 02n =−−− hRFFMg ll

Solve for Fn: ( )l

hRFMgF −−=

2n

Express l as a function of R and h: ( ) 222 2 hRhhRR −=−−=l

( )

hhRFMg

hRhhRFMgF

−−=

−

−−=

2

22

2n

(b) Apply ∑ = 0xF to the cylinder: 0h,c =+− FF

Solve for Fc,h: FF =h,c

(c) Apply∑ = 0yF to the cylinder: 0v,cn =+− FMgF

Solve for Fc,v: nvc, FMgF −=

Chapter 12

932

Substitute the result from part (a):

hhRF

hhRFMgMgF

−=

⎭⎬⎫

⎩⎨⎧ −

−−=

2

2vc,

36 •• Picture the Problem The figure to the right shows the forces acting on the cylinder. Because the cylinder is in equilibrium, we can use the condition for rotational equilibrium to express Fn in terms of F. Because, to roll over the step, the cylinder must lift off the floor, we can set Fn = 0 in our expression relating Fn and F and solve for F.

Apply 0=∑τr about the step’s corner:

( ) 02n =−−− hRFFMg ll

Solve for Fn: ( )l

hRFMgF −−=

2n

Express l as a function of R and h: ( ) 222 2 hRhhRR −=−−=l


hhRFMg

hRhhRFMgF

−−=

−

−−=

22

22n

To roll over the step, the cylinder must lift off the floor, i.e., Fn = 0: h

hRFMg −−=

20

Solve for F:

hRhMgF−

=2

*37 •• Picture the Problem The diagram shows the forces F1 and F2 that the fencer’s hand exerts on the epee. We can use a condition for translational equilibrium to find the upward force the fencer must exert on the epee when it is in equilibrium and the definition of torque to determine the total torque exerted. In part (c) we can use the conditions for translational and rotational equilibrium to obtain two equations in F1 and


933

F2 that we can solve simultaneously. In part (d) we can apply Newton’s 2nd law in rotational form and the condition for translational equilibrium to obtain two equations in F1 and F2 that, again, we can solve simultaneously.

(a) Letting the upward force exerted by the fencer’s hand be F, apply 0=∑ yF to the epee to obtain:

0=−WF

Solve for and evaluate F:

( )( ) N87.6m/s81.9kg7.0 2 === mgF

(b) Express the torque due to the weight about the left end of the epee:

( )( ) mN65.1N87.6m24.0 ⋅=== wlτ

(c) Apply 0=∑ yF to the epee to obtain:

0N87.621 =−+− FF (1)

Apply 00=∑τ to obtain:

( ) ( ) 0mN65.1m12.0m02.0 21 =⋅−+− FF

Solve these equations simultaneously to obtain:

N26.81 =F and N1.152 =F .

Note that the force nearest the butt of the epee is directed downward and the force nearest the hand guard is directed upward.

38 •• Picture the Problem In the force diagram, the forces exerted by the hinges are ,2,yF

r,1,yF

r and 1,xF

rwhere the subscript 1 refers to the lower hinge. Because the gate

is in equilibrium, we can apply the conditions for translational and rotational equilibrium to find the tension in the wire and the forces at the hinges.

Chapter 12

934


through the lower hinge and perpendicular to the plane of the page:

0cossin 121 =−+ mgTT lll θθ

Solve for T: θθ cossin 21

1

ll

l

+=

mgT


( )( )( ) ( )

N141

45cosm5.145sinm5.1N200m5.1

=

°+°=T

(b) Apply ∑ = 0xF to the gate: 045cos1, =°−TFx

Solve for and evaluate Fx,1: ( )

N7.99

45cosN14145cos1,

=

°=°= TFx

(c) Apply ∑ = 0yF to the gate: 045sin2,1, =−°++ mgTFF yy

Because Fy,1 and Fy,2 cannot be determined independently, solve for and evaluate their sum: N100

N99.7N200

45sin2,1,

=

−=

°−=+ TmgFF yy

39 ••• Picture the Problem Let T = the tension in the wire; Fn = the normal force of the surface; and fs,max = µsFn the maximum force of static friction. Letting the point at which


935

the wire is attached to the log be the origin, the center of mass of the log is at (−1.838 m, −0.797 m) and the point of contact with the floor is at (−3.676 m, −1.594 m). Because the log is in equilibrium, we can apply the conditions for translational and rotational equilibrium.

Apply ∑ = 0xF to the log: 0sin maxs, =− fT θ

or nsmaxs,sin FfT µθ == (1)

Apply ∑ = 0yF to the log: 0cos n =−+ mgFT θ

or ncos FmgT −=θ (2)

Divide equation (1) by equation (2) to obtain: n

ns

cossin

FmgF

TT

−=

µθθ

or

1tan

n

s1

−= −

Fmgµθ (3)


through the origin:

0ns3n12 =−− FFmg µlll

Solve for Fn: s31

2n µll

l

+=

mgF

Chapter 12

936

Substitute numerical values and evaluate Fn:

( )( )( ) N3890.61.5943.676

m/s9.81kg1001.838 2

n =+

=F

Substitute in equation (3) and evaluate θ: ( )( )

°=

−= −

5.21

1N893

m/s9.81kg1006.0tan 2

1θ

Substitute numerical values in equation (1) and evaluate T:

( )( ) N636sin21.5

N3896.0=

°=T

40 ••• Picture the Problem Consider what happens just as θ increases beyond θ max. Because the top of the block is fixed by the cord, the block will in fact rotate with only the lower right edge of the block remaining in contact with the plane. It follows that just prior to this slipping, Fn and fs = µsFn act at the lower right edge of the block. Choose a coordinate system in which up the incline is the positive x direction and the direction of nF

ris the

positive y direction. Because the block is in equilibrium, we can apply the conditions for translational and rotational equilibrium.

Apply ∑ = 0xF to the block: 0sinns =−+ θµ mgFT (1)

Apply ∑ = 0yF to the block: 0cosn =− θmgF (2)


through the lower right edge of the block:

( ) ( ) 0sincos 21

21 =−+ bTmgbmga θθ (3)

Eliminate Fn between equations (1) and (2) and solve for T:

( )θµθ cossin s−= mgT


937

Substitute for T in equation (3): ( ) ( )( )[ ] 0cossin

sincos

s

21

21

=−−

+

θµθθθ

mgbmgbmga

Substitute 4a for b: ( ) ( )( )

( ) ( )[ ] 0cossin4sin4cos

s

21

21

=−−+

θµθθθ

mgamgamga

Simplify to obtain: ( ) 0sin4cos81 s =−+ θθµ

Solve for θ: 4

81tan s1 µθ += −


( )°=

+= − 6.61

48.081tan 1θ

*41 •• Picture the Problem The free-body diagram shown to the left below is for the weight and the diagram to the right is for the boat. Because both are in equilibrium under the influences of the forces acting on them, we can apply a condition for translational equilibrium to find the tension in the chain.

(a) Apply 0=∑ xF to the boat: 0cosd =− θTF

Solve for T: θcos

dFT =

Apply 0=∑ yF to the weight: 0N100sin2 =−θT (1)

Substitute for T to obtain:

0N100tan2 d =−θF

Solve for θ :

d

1

2N100tan

F−=θ

Chapter 12

938

Substitute for Fd and evaluate θ : ( ) °== − 45

N502N100tan 1θ

Solve equation (1) for T: θsin2

N100=T

Substitute for θ and evaluate T: N7.70

45sin2N100

=°

=T

(b) Use the diagram to the right to relate the sag ∆y in the chain to the angle θ the chain makes with the horizontal:

Ly

21

sin ∆=θ

where L is the length of the chain.

Solve for ∆y:

θsin21 Ly =∆

Because the horizontal and vertical forces in the chain are equal, θ = 45° and:

( ) m77.145sinm521 =°=∆y

(c) Relate the distance d of the boat from the dock to the angle θ the chain makes with the horizontal:

Ld

Ld==

2121

cosθ

Solve for and evaluate d: ( ) m54.345cosm5cos =°== θLd

(d) Relate the resultant tension in the chain to the vertical component of the tension Fv and the maximum drag force exerted on the boat by the water Fd,max:

( )22maxd,

2v N500=+ FF

Solve for Fd,max: ( ) 2v

2maxd, N500 FF −=

Because the vertical component of the tension is 50 N: ( ) ( ) N497N50N500 22

maxd, =−=F

42 •• Picture the Problem Choose a coordinate system in which the positive x axis is along the rod and the positive y direction is normal to the rod. The rod and the forces acting on


939

it are shown in the free-body diagram. The forces acting at the supports are denoted by the numerals 1 and 2. The resultant forces at the supports are shown as dashed lines. We’ll assume that the rod is on the verge of sliding. Because the x components of the forces at the supports are friction forces, they are proportional to the normal, i.e., y, components of the forces at the supports. Because the rod is in equilibrium, we can apply the conditions for translational and rotational equilibrium.


through the support at x = 2 m:

0cos1,22 =− θmgF y ll

Solve for F2,y:

2

1,2

cosl

l θmgF y =

Substitute numerical values and evaluate F2,y:

( )( )( )

N4.127m4

30cosm/s9.81kg20m3 2

,2

=

°=yF


through the support at x = 6 m:

( ) 0cos ,1212 =−− yFmg lll θ

Solve for F1,y:

( )2

12,1

cosl

ll θmgF y−

=

Substitute numerical values and evaluate F1,y:

( )( )( )

N48.4230cos

m4m/s9.81kg20m3m4 2

,1

=°×

−=yF

Chapter 12

940

Apply ∑ = 0xF to the rail: 030sin,2,1 =°−+ mgFF xx (1)

Assuming that the rod is on the verge of sliding and that the coefficient of static friction is the same for both supports:

yFF ,1sx1, µ=

and yFF ,2sx2, µ=

Divide the first of these equations by the second and evaluate this ratio to obtain:

31

N127.4N48.42

,2

,1

,2

,1 ===y

y

x

x

FF

FF

Solve for F2,x: xx FF ,1,2 3=

Substitute in equation (1): 0sin3 ,1,1 =−+ θmgFF xx

Solve for F1,x: θsin4

1,1 mgF x =

Substitute numerical values and evaluate F1,x:

( )( )N53.24

30sinm/s9.81kg20 241

,1

=

°=xF

Evaluate F2,x: ( ) N58.73N53.243,2 ==xF

Find the angle θ1 the force at support 1 (x = 2 m) makes with the rod:

°=== −− 0.60N53.24N48.42tantan 1

,1

,111

x

y

FF

θ

Find the angle θ2 the force at support 2 makes with the rod:

°=== −− 0.60N58.73N4.127tantan 1

,2

,212

x

y

FF

θ

Find the magnitude of 1Fr

:

( ) ( )N49.1

N42.48N24.53 22

2,1

2,11

=

+=

+= yx FFF


941

Find the magnitude of 2Fr

:

( ) ( )N471

N4.271N58.37 22

2,2

2,22

=

+=

+= yx FFF

43 • Picture the Problem The forces shown in the figure constitute a couple and will cause the plate to experience a counterclockwise angular acceleration. We can find this net torque by expressing the torque about either of the corners of the plate. Sum the torques about an axis through the upper left corner of the plate to obtain:

( )[ ] ( )[ ]( ) ( )ab

ab

N0.40N3.69

30sinN8030cosN80net

−=

°−°=τ

44 • Picture the Problem We can use the condition for translational equilibrium and the definition of a couple to show that the force of static friction exerted by the surface and the applied force constitute a couple. We can use the definition of torque to find the torque exerted by the couple. We can use our result from (b) to find the effective point of application of the normal force when F = Mg/3 and the condition for rotational equilibrium to find the greatest magnitude of F

r for which the cube will not tip.

(a) Apply ∑ = 0xF

rto the stationary

cube:

0s =+ fFrr

couple. a constitute forces directed oppositely and parallel,

equal, ofpair thisand sfFrr

−=∴

The torque of the couple is: Fa=coupleτ

(b) Let x = the distance from the point of application of Fn to the center of the cube. Now, Fn = Mg, so applying 0=∑τr to the cube

yields:

0=− FaMgx (1)

or

MgFax =

Substitute for F = Mg/3 to obtain:

33 aMg

aMg

x ==

Chapter 12

942

(c) Solve equation (1) for F: a

MgxF =

Noting that xmax = a/2, substitute to express the condition that the cube will tip: 2

2max Mga

aMg

aMgxF ==>

45 •• Picture the Problem We can find the perpendicular distance between the lines of action of the two forces by following the outline given in the problem statement. Express the vertical components of the forces: FF

2330cos =°

Express the horizontal components of the forces: 2

30sin FF =°

Express the net torque acting on the plate: ( )abFFaFb −=−= 3

21

21

23

netτ

Letting D be the moment arm of the couple, express the net torque acting on the plate:

FD=netτ

Equate these two expressions for τnet: ( )abFFD −= 321

Solve for D: ( )abD −= 32

1

*46 •• Picture the Problem Choose the coordinate system shown in the diagram and let x be the coordinate of the thrust point. The diagram to the right shows the forces acting on the wall. The normal force must balance out the weight of the wall and the vertical component of the thrust from the arch and the frictional force must balance out the horizontal component of the thrust. We can apply the conditions for translational equilibrium to find f and Fn and the condition for rotational equilibrium to find the distance x from the origin of our coordinate system at which Fn acts.


943

(a) Apply the conditions for translational equilibrium to the wall to obtain:

∑ =+−= 0cosarch θFfFx (1) and

∑ =−−= 0sinarchn θFmgFFy (2)

Solve equation (1) for and evaluate f: ( )kN3.17

30cosN102cos 4arch

=

°×== θFf

Solve equation (2) for Fn: θsinarchn FmgF +=


( )( )( )

kN304

30sinN102

m/s81.9kg1034

24n

=

°×+

×=F

Apply 0axis =∑ zτ to the to the wall: 0cosarch2

1n =−− θhFwmgxF

Solve for x:

n

arch21 cos

FhFwmgx θ+

=


( )( )( ) ( )( ) m570.0kN304

30cosN102m10m/s81.9kg103m25.1 42421

=°×+×

=x

(b)

center. thecloser to force normal theofaction ofpoint themovemust thrust the tocomparedlarger weight themaking so mass, ofcenter ethrough th

act wouldforce normal the wall, theof side on the thrust no were thereIf

47 •• Picture the Problem Let h be the height of the structure, T be the thrust, θ the angle from the horizontal of the thrust, m′g the weight of the wall above height y, N(x) the normal force, f the friction force the lower part of the wall exerts on the upper part, and w the width of the structure. We can apply the conditions for translational and rotational equilibrium to the portion of the wall above the point at which the thrust is applied to obtain two equations that we can solve simultaneously for x.

Chapter 12

944

Apply 0=∑ yF to that fraction of the wall above height y:

( ) 0sin =−− m'gTxN θ

Assuming the wall is of uniform density, express m′g in terms of mg: h

mgyh

m'g=

−

and

⎟⎠⎞

⎜⎝⎛ −=

hymgm'g 1

Substitute to obtain: ( ) 01sin =⎟

⎠⎞

⎜⎝⎛ −−−

hymgTxN θ

Solve for N(x): ( ) ⎟

⎠⎞

⎜⎝⎛ −+=

hymgTxN 1sinθ

Apply 0=∑τr about an axis through (0,y) and perpendicular to the xy plane to obtain:

( ) ( )

01

cos

21 =⎟

⎠⎞

⎜⎝⎛ −−

−−

hymgw

TyhxxN θ


( ) ⎟⎠⎞

⎜⎝⎛ −

+=

hy

xNhTmgwx 1cos2

1 θ

Substitute for N(x) to obtain:

( )

⎟⎠⎞

⎜⎝⎛ −+

⎟⎠⎞

⎜⎝⎛ −+

=

hymgT

hyhTmgw

x1sin

1cos21

θ

θ


( )( )( ) ( )( )[ ]

( ) ( )( )

( )yy

y

y

x

1-

244

42421

m943.243.30571.3m71.35

m101m/s81.9kg10330sinN102

m10130cosN102m10m25.1m/s81.9kg103

−−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−×+°×

⎟⎟⎠

⎞⎜⎜⎝

⎛−°×+×

=

Solve for y: ( )x

xy 1-m943.2571.343.30m71.35

−−

=


945

The graph shown below was plotted using a spreadsheet program:

0

2

4

6

8

10

12

0.0 0.2 0.4 0.6 0.8 1.0 1.2

x (m)

y (m

)

Ladder Problems *48 •• Picture the Problem The ladder and the forces acting on it at the critical moment of slipping are shown in the diagram. Use the coordinate system shown. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium.

Using its definition, express µs:

n

maxs,s F

f=µ (1)

Apply 0=∑τr about the bottom of

the ladder:

( )[ ] ( )[ ]( )[ ] 0sinm10

cosm5cosm9

W =−+

FmgMg

θθθ

Solve for FW: ( ) ( )

( ) θθ

cossinm10

m5m9W gmMF +=

Find the angle θ: °== − 74.73

m10m2.8cos 1θ

Chapter 12

946

Evaluate FW: ( )( ) ( )( )( )

( )N7.211

74.73cosm/s9.8174.73sinm10

kg22m5kg70m9

2

W

=°×

°+

=F

Apply ∑ = 0xF to the ladder and

solve for fs,max:

0maxs,W =− fF

and N7.211Wmaxs, == Ff

Apply ∑ = 0yF to the ladder: 0n =−− mgMgF

Solve for Fn: ( )

( )( )N5.902

m/s9.81kg22kg70 2n

=+=

+= gmMF

Substitute numerical values in equation (1) and evaluate µs:

235.0N902.5N211.7

s ==µ

49 •• Picture the Problem The ladder and the forces acting on it are shown in the diagram. Because the wall is smooth, the force the wall exerts on the ladder must be horizontal. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium to it.

Apply ∑ = 0yF to the ladder and

solve for Fn:

0n =−MgF ⇒ MgF =n


solve for fs,max:

0maxs,W =− fF ⇒ Wmaxs, Ff =

Apply 0=∑τr about the bottom of

the ladder:

0sincos W =− θθ LFMgx


947

Solve for x:

θµθµ

θθθ

tantan

tancossin

sns

maxs,W

LMg

LFMg

LfMg

LFx

==

==

Referring to the figure, relate x to h and solve for h: x

h=θsin

and θθµθ sintansin sLxh ==

50 •• Picture the Problem The ladder and the forces acting on it are shown in the drawing. Choose a coordinate system in which the positive x direction is to the right and the positive y direction is upward. Because the wall is smooth, the force the wall exerts on the ladder must be horizontal. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium. Apply ∑ = 0yF to the ladder and

solve for Fn:

04n =−− mgmgF

and mgF 5n =


solve for fs,max:

0maxs,W =− fF

and Wmaxs, Ff =


through the bottom of the ladder:

0sincos4cos2 W =−+ θθθ LFmgLmg l

Substitute for FW and then fs,max and solve for l: θ

θθµcos4

cossin5 21

s

mgmgLmgL −

=l

Chapter 12

948

Simplify to obtain:

( )

L

L

L

849.0

8160tan

445.05

81tan

45 s

=

⎟⎠⎞

⎜⎝⎛ −°=

⎟⎠⎞

⎜⎝⎛ −= θµ

l

i.e., you can climb about 85% of the way to the top of the ladder.

51 •• Picture the Problem The ladder and the forces acting on it are shown in the figure. Because the ladder is separating from the wall, the force the wall exerts on the ladder is zero. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium.

To find the force required to pull the ladder away from the wall, apply

0=∑τr about an axis through the

bottom of the ladder:

0sin2

cos2

=− θθ FLLmg

or, because θ

θtan

cos2

hL= ,

0sin2tan

=− θθ

FLmgh

Solve for F:

θθ sintan2

LmghF = (1)

Apply ∑ = 0xF to the ladder: nsmaxs,maxs, 0 FfFfF µ==⇒=− (2)

Apply ∑ = 0yF to the ladder: mgFmgF =⇒=− nn 0

Equate equations (1) and (2) and substitute for Fn to obtain: θθ

µsintan

2s L

mghmg =

Solve for µs:

θθµ

sintan2

s Lh

=


949

52 •• Picture the Problem Assume that half the man’s weight acts on each side of the ladder. The force exerted by the frictionless floor must be vertical. D is the separation between the legs at the bottom and x is the distance of the cross brace from the apex. Because each leg of the ladder is in equilibrium, we can apply the condition for rotational equilibrium the right leg to relate the tension in the cross brace to its distance from the apex. (a) By symmetry, each leg carries half the total weight. So the force on each leg is:

N450

(b) Consider one of the ladder’s legs and apply 0=∑τr about the

apex:

02n =−TxDF

Solve for T: xDFT

2n=

Using trigonometry, relate h and θ through the tangent function:

hD 2tan 2

1 =θ

Solve for D to obtain:

θ21tan2hD =

Substitute and simplify to obtain: x

hFx

hFT θθ 21

n21

n tan2tan2

==


xhFT θ2

1n tan

=

Apply 0=∑ yF to the ladder and

solve for Fn:

021

n =− wF and wF 21

n =

Substitute to obtain: x

whT2tan 2

1θ= (1)

Chapter 12

950


( )( )( ) N241

m2215tanm4N900

=°

=T

(c) From equation (1) we can see that, if x is increased, i.e., the brace moved lower:

decrease. willT

53 •• Picture the Problem The figure shows the forces acting on the ladder. Because the wall is frictionless, the force the wall exerts on the ladder is perpendicular to the wall. Because the ladder is on the verge of slipping, the static friction force is fs,max. Because the ladder is in equilibrium, we can apply the conditions for translational and rotational equilibrium. Apply ∑ = 0xF to the ladder:

nsmaxs,Wmaxs,W 0 FfFfF µ==⇒=−

Apply ∑ = 0yF to the ladder:

mgFmgF =⇒=− nn 0



0sincos2 W =− θθ LFLmg

Substitute for FW and Fn and simplify to obtain:

0sincos s21 =− θµθ

Solve for and evaluate θ: ( ) °=== −− 0.59

3.021tan

21tan 1

s

1

µθ

Stress and Strain *54 • Picture the Problem L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. The stretch in the wire ∆L is related to Young’s modulus by

( ) ( ).LLAFY ∆= We can use Table 12-1 to find the numerical value of Young’s

modulus for steel. Find the amount the wire is stretched from Young’s modulus: LL

AFY∆

=


951

Solve for ∆L:

YAFLL =∆

Substitute for F and A to obtain:

2rYmgLLπ

=∆

Substitute numerical values and evaluate ∆L:

( )( )( )( )

mm976.0

m102N/m102m5m/s9.81kg50

23211

2

=

××=∆

−πL

55 • Picture the Problem L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. The stretch in the wire ∆L is related to Young’s modulus by

( ) ( ).strainstress LLAFY ∆==

(a) Express the maximum load in terms of the wire’s breaking stress: 2

max

stressbreaking

stressbreaking

r

AF

π×=

×=

Substitute numerical values and evaluate Fmax:

( ) ( )N41.6

m100.21N/m103 2328max

=

××= −πF

(b) Using the definition of Young’s modulus, express the fractional change in length of the copper wire: %136.01036.1

N/m101.1N/m101.5

3

211

28

=×=

××

==∆

−

YAFLL

56 • Picture the Problem L is the unstretched length of the wire, F is the force acting on it, and A is its cross-sectional area. The stretch in the wire ∆L is related to Young’s modulus by

( ) ( ).LLAFY ∆= We can use Table 12-1 to find the numerical value of Young’s

modulus for steel. Find the amount the wire is stretched from Young’s modulus: LL

AFY∆

=

Solve for ∆L:

YAFLL =∆

Chapter 12

952

Substitute for F and A to obtain: 2rY

mgLLπ

=∆


( )( )( )( )

mm833.0

m103.0N/m102m2.1m/s9.81kg4

23211

2

=

××=∆

−πL

*57 • Picture the Problem The shear stress, defined as the ratio of the shearing force to the area over which it is applied, is related to the shear strain through the definition of the shear

modulus; θtanstrainshear

stressshear ss

AFM == .

Using the definition of shear modulus, relate the angle of shear, θ to the shear force and shear modulus:

AMF

s

stan =θ

Solve for θ : AM

F

s

s1tan−=θ

Substitute numerical values and evaluate θ : ( )( )

°=

××= −

−

01.5

m1015N/m101.9N25tan 2425

1θ

58 •• Picture the Problem The stretch in the wire ∆L is related to Young’s modulus by ( ) ( )LLAFY ∆= , where L is the unstretched length of the wire, F is the force acting

on it, and A is its cross-sectional area. For a composite wire, the length under stress is the unstressed length plus the sum of the elongations of the components of the wire. Express the length of the composite wire when it is supporting a mass of 5 kg:

LL ∆+= m00.3 (1)

Express the change in length of the composite wire:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

∆+∆=∆

Al

Al

steel

steel

Al

Al

steel

steel

Alsteel

YL

YL

AF

YL

AF

YL

AF

LLL


953

Find the stress in each wire: ( )( )

( )27

23

2

N/m10245.6m100.5

m/s9.81kg5

×=

×=

−πAF


( ) m1081.1N/m107.0

m1.5N/m102m1.5N/m10245.6 3

21121127 −×=⎟⎟

⎠

⎞⎜⎜⎝

⎛×

+×

×=∆L

Substitute in equation (1) and evaluate L:

m0018.3

m1081.1m00.3 3

=

×+= −L

59 •• Picture the Problem We can use Hooke’s law and Young’s modulus to show that, if the wire is considered to be a spring, the force constant k is given by k = AY/L. By treating the wire as a spring we can show the energy stored in the wire is U = ½F∆L. Express the relationship between the stretching force, the stiffness constant, and the elongation of a spring:

LkF ∆= or

LFk∆

=

Using the definition of Young’s modulus, express the ratio of the stretching force to the elongation of the wire:

LAY

LF

=∆

(1)

Equate these two expressions for F/∆L to obtain:

LAYk =

Treating the wire as a spring, express its stored energy:

( ) ( ) 2212

21 L

LAYLkU ∆=∆=

Solve equation (1) for F:

LLAYF ∆

=

Substitute in our expression for U to obtain: LFL

LLAYU ∆=∆

∆= 2

121

Chapter 12

954

60 •• Picture the Problem Let L′ represent the stretched and L the unstretched length of the wire. The stretch in the wire ∆L is related to Young’s modulus by ( ) ( )LLAFY ∆= , where F is

the force acting on it, and A is its cross-sectional area. In problem 58 we showed that the energy stored in the wire is U = ½F∆L, where Y is Young’s modulus and ∆L is the amount the wire has stretched. (a) Express the stretched length of the wire:

LLL' ∆+=

Using the definition of Young’s modulus, express ∆L: AY

LFL =∆

Substitute and simplify:

⎟⎠⎞

⎜⎝⎛ +=+=

AYFL

AYLFLL' 1

Solve for L:

AYF

L'L+

=1


( ) ( )m347.0

N/m102m100.1N531

m0.35

21123

=

××+

=

−π

L

(b) Using the expression from Problem 59, express the work done in stretching the wire:

( )( )J0.0795

m0.347m0.35N5321

21

=

−=∆=∆= LFUW

*61 •• Picture the Problem The table to the right summarizes the ratios ∆L/F for the student’s data. Note that this ratio is constant, to three significant figures, for loads less than or equal to 200 g. We can use this ratio to calculate Young’s modulus for the rubber strip.

Load F ∆L ∆L/F (g) (N) (m) (m/N 100 0.981 0.006 6.12×10−3 200 1.962 0.012 6.12×10−3 300 2.943 0.019 6.46×10−3 400 3.924 0.028 7.14×10−3 500 4.905 0.05 10.2×10−3


955

(a) Referring to the table, we see that for loads ≤ 200 g:

m/N1012.6 3−×=∆FL

Use the definition of Young’s modulus to express Y:

FLA

LLA

FLY∆

=∆

=

Substitute numerical values and evaluate Y:

( ) ( ) ( )26

333

2

N/m1082.1m/N106.12m101.5m103

m105×=

××××

= −−−

−

Y

(b) Interpolate to determine the stretch when the load is 150 g, and use the expression from Problem 58, to express the energy stored in the strip:

( )( )( )mJ62.6

m109m/s81.9kg0.15 3221

21

=

×=

∆=−

LFU

62 •• Picture the Problem The figure shows the forces acting on the wire where it passes over the nail. m represents the mass of the mirror and T is the tension in the supporting wires. The figure also shows the geometry of the right triangle defined by the support wires and the top of the mirror frame. The distance a is fixed by the geometry while h and L will change as the mirror is suspended from the nail. Express the distance between the nail and the top of the frame when the wire is under tension:

hhhh'∆+=

∆+=m4.0

(1)

Apply ∑ = 0yF to the wire where it

passes over the supporting nail:

0cos2 =− θTmg

Solve for the tension in the wire: θcos2

mgT =

Chapter 12

956


( )( ) N0.25

m0.85m0.42

m/s9.81kg2.4 2

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=T

Using its definition, find the stress in the wire: ( )

28

23

N/m1096.7m101.0

N0.25stress

×=

×==

−πAT

Using the definition of Young’s modulus, find the strain in the hypotenuse of the right triangle shown in the figure:

3211

28

1098.3N/m102

N/m1096.7

stressstrain

−×=××

=

=∆

=YL

L

Using the Pythagorean theorem, express the relationship between the sides of the right triangle in the figure:

222 Lha =+

Express the differential of this equation:

LLhhaa ∆=∆+∆ 222 or, because ∆a = 0,

LLhh ∆=∆

Solve for and evaluate ∆h: LL

hL

hLLh ∆

⋅=∆

=∆2

Substitute numerical values and evaluate ∆h:

( ) ( ) mm 7.19103.98m0.4m0.85 3

2

=×=∆ −h


cm72.40

mm19.7m4.0

=

+=h'

63 •• Picture the Problem Let the numeral 1 denote the aluminum wire and the numeral 2 the steel wire. Because their initial lengths and amount they stretch are the same, we can use the definition of Young’s modulus to express the change in the lengths of each wire and then equate these expressions to obtain an equation solvable for the ratio M1/M2. Using the definition of Young’s modulus, express the change in Al1

111 YA

gLML =∆


957

length of the aluminum wire: Using the definition of Young’s modulus, express the change in length of the steel wire:

steel2

222 YA

gLML =∆

Because the two wires stretch by the same amount, equate ∆L1 and ∆L2 and simplify:

steel2

2

Al1

1

YAM

YAM

=

Solve for the ratio M1/M2:

steel2

Al1

2

1

YAYA

MM

=

Substitute numerical values and evaluate M1/M2:

( ) ( )( ) ( )

( ) ( )( ) ( )

686.0

N/m102mm5.0N/m107.0mm7.0

N/m102mm5.04

N/m107.0mm7.04

2112

2112

2112

2112

2

1

=

××

=

×

×= π

π

MM

64 •• Picture the Problem The free-body diagram shows the forces acting on the ball as it rotates around the post in a horizontal plane. We can apply Newton’s 2nd law to find the tension in the wire and use the definition of Young’s modulus to find the amount by which the aluminum wire stretches.

Express the length of the wire under tension to its unstretched length:

LLLL ∆+=∆+= m7.00 (1)

Apply ∑ = 0yF to the ball: 0sin =−mgT θ

Solve for the tension in the wire:

θsinmgT =

Chapter 12

958


( )( ) N3.56sin5

m/s9.81kg0.5 2

=°

=T

Using the definition of Young’s modulus, express ∆L: AY

FLL =∆


( )( )( ) ( )

mm280.0

N/m107.0m106.14

m0.7N56.321123

=

××=∆

−πL

Substitute in equation (1) to obtain: cm03.70mm280.0m7.0 =+=L

*65 •• Picture the Problem We can use the definition of stress to calculate the failing stress of the cable and the stress on the elevator cable. Note that the failing stress of the composite cable is the same as the failing stress of the test sample. Express the stress on the elevator cable:

210

26cable

N/m1067.1m102.1

kN20Stress

×=

×== −A

F

Express the failing stress of the sample:

210

26failing

N/m10500.0m102.0

kN1Stress

×=

×== −A

F

elevator. esupport thnot it will ,StressStress Because cablefailing <

*66 ••• Picture the Problem Let the length of the sides of the rectangle be x, y and z. Then the volume of the rectangle will be V = xyz and we can express the new volume V ′ resulting from the pulling in the x direction and the change in volume ∆V in terms of ∆x, ∆y, and ∆z. Discarding the higher order terms in ∆V and dividing our equation by V and using the given condition that ∆y/y = ∆z/z will lead us to the given expression for ∆y/y. Express the new volume of the rectangular box when its sides change in length by ∆x, ∆y, and ∆z:

( )( )( ) ( ) ( ) ( ) zyxzyxzxyyxz

xyzxzyyzxxyzzzyyxxV'∆∆∆+∆∆+∆∆+∆∆+

∆+∆+∆+=∆+∆+∆+=

where the terms in brackets are very small (i.e., second order or higher).


959

Discard the second order and higher terms to obtain:

( ) ( ) ( )xyzxzyyzxVV' ∆+∆+∆+=

or ( ) ( ) ( )xyzxzyyzxVV'V ∆+∆+∆=−=∆

Because ∆V = 0: ( ) ( ) ( )[ ]xyzxzyyzx ∆+∆−=∆

Divide both sides of this equation by V = xyz to obtain:

⎥⎦

⎤⎢⎣

⎡ ∆+

∆−=

∆zz

yy

xx

Because ∆y/y = ∆z/z, our equation becomes: y

yxx ∆

−=∆ 2 or

xx

yy ∆

−=∆

21

67 •• Picture the Problem We can evaluate the differential of the volume of the wire and, using the assumptions that the volume of the wire does not change under stretching and that the change in its length is small compared to its length, show that ∆r/r = −(1/2) ∆L/L. Express the volume of the wire:

LrV 2π=

Evaluate the differential of V to obtain:

rLdrdLrdV ππ 22 +=

Because dV = 0: LdrrdL 20 += ⇒

LdL

rdr

21

−=

Because ∆L << L, we can approximate the differential changes dr and dL with small changes ∆r and ∆L to obtain:

LL

rr ∆

−=∆

21

*68 ••• Picture the Problem Because the volume of the thread remains constant during the stretching process, we can equate the initial and final volumes to express r0 in terms of r. We can also use Young’s modulus to express the tension needed to break the thread in terms of Y and r0. (a) Express the conservation of volume during the stretching of the spider’s silk:

02

02 LrLr ππ =

Solve for r:

LLrr 0

0=

Substitute for L to obtain:

00

00 316.0

10r

LLrr ==

Chapter 12

960

(b) Express Young’s modulus in terms of the breaking tension T: LL

rTLLrT

LLATY

∆=

∆=

∆=

20

2 10 ππ

Solve for T to obtain: L

LYrT ∆= 2

0101 π

Because ∆L/L = 9:

109 2

0 YrT π=

General Problems 69 • Picture the Problem Because the board is in equilibrium, we can apply the conditions for translational and rotational equilibrium to determine the forces exerted by the supports. Apply 0=∑i iτ

rabout the right support:

( )( ) ( )( ) ( ) 0m10N90m5N360m2 L =−+ F

Solve for and evaluate FL: ( )( ) ( )( )

N117

m10N90m5N360m2

L

=

+=F

Apply ∑ = 0yF to the board: 0N360N90RL =−−+ FF

Solve for and evaluate FR:

N333

N360N90N117N360N90LR

=

++−=++−= FF

Remarks: We could just as easily found FR by applying 0=∑τr about the left

support. 70 • Picture the Problem Because the man-and-board system is in equilibrium, we can apply the conditions for translational and rotational equilibrium to determine the forces exerted by the supports. Let d represent the distance from the man’s feet to his center of gravity. Apply 0=∑τr about an axis

through the man’s feet and perpendicular to the page:

( ) ( )( ) 0N454m88.1N458 =−d


961

Solve for and evaluate d: ( )( )

cm0.99

m990.0N845

N454m88.1

=

==d

change.not wouldreadings scale theso and mass ofcenter hisoflocation thechangenot wouldboard theaboveslightly head his Holding No.

*71 • Picture the Problem We can apply the balance condition 0=∑τr successively, starting

with the lowest part of the mobile, to find the value of each of the unknown weights. Apply 0=∑τr about an axis

through the point of suspension of the lowest part of the mobile:

( )( ) ( ) 0cm4N2cm3 1 =− w

Solve for and evaluate w1: ( )( ) N50.1cm4

N2cm31 ==w


through the point of suspension of the middle part of the mobile:

( ) ( )( ) 0N1.5N2cm4cm2 2 =+−w


N1.5N2cm42 =

+=w


through the point of suspension of the top part of the mobile:

( )( ) ( ) 0cm6N5.10cm2 3 =− w


N5.10cm23 ==w

72 • Picture the Problem We can determine the ratio of L to h by noting the number of ropes supporting the load whose mass is M. (a) Noting that three ropes support the pulley to which the object whose mass is M is fastened we can

3=hL

Chapter 12

962

conclude that: (b) Apply the work-energy principle to the block-tackle object to obtain:

tackle-blocksystemext UEW ∆=∆=

or MghFL =

73 •• Picture the Problem The figure shows the equilateral triangle without the mass m, and then the same triangle with the mass m and rotated through an angle θ. Let the side length of the triangle to be 2a. Then the center of mass of the triangle is at a

distance of 3

2a from each vertex. As the

triangle rotates, its center of mass shifts

by3

2aθ, for θ << 1. Also, the vertex to

which m is attached moves toward the plumb line by the distance d = 2aθ cos30° = θa3 (see the drawing).


the point of suspension:

( ) 03

23 =−− θθ aMgaamg

Solve for m/M: ( )θθ

3132−

=Mm

Substitute numerical values and evaluate m/M:

( )

( )

148.0

180rad6313

180rad62

=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

°°−

⎟⎠⎞

⎜⎝⎛

°°

=π

π

Mm


963

74 •• Picture the Problem If the hexagon is to roll rather than slide, the incline’s angle must be such that the center of mass falls just beyond the support base. From the geometry of the hexagon, one can see that the critical angle is 30°. The free-body diagram shows the forces acting on the hexagonal pencil when it is on the verge of sliding. We can use Newton’s 2nd law to relate the coefficient of static friction to the angle of the incline for which rolling rather than sliding occurs.

Apply ∑ = 0F

rto the pencil: 0sin maxs, =−=∑ fmgFx θ (1)

and 0cosn =−=∑ θmgFFy (2)

Substitute fs,max = µsFn in equation (1): 0sin ns =− Fmg µθ (3)


stan µθ =

Thus, if the pencil is to roll rather than slide when the pad is inclined:

577.030tans =°≥µ

75 •• Picture the Problem The box and the forces acting on it are shown in the figure. When the box is about to tip, Fn acts at its edge, as indicated in the drawing. We can use the definition of µs and apply the condition for rotational equilibrium in an accelerated frame to relate fs to the weight of the box and, hence, to the normal force. Using its definition, express µs:

n

ss F

f≥µ

Chapter 12

964


the box’s center of mass:

0n21

s =− wFwf

Solve for the ratio n

s

Ff

: 21

n

s =Ff

Substitute to obtain the condition for tipping:

500.0s ≥µ

Therefore, if the box is to slide: 500.0s <µ

76 •• Picture the Problem Because the balance is in equilibrium, we can use the condition for rotational equilibrium to relate the masses of the blocks to the lever arms of the balance in the two configurations described in the problem statement. Apply 0=∑τr about an axis

through the fulcrum:

( ) ( ) 21 kg95.1kg5.1 LL =

Solve for the ratio L1/L2: 30.1kg5.1kg95.1

2

1 ==LL


through the fulcrum with 1.5 kg at L2:

( ) 12kg5.1 MLL =

Solve for and evaluate M: ( )

kg15.1

30.1kg5.1kg5.1kg5.1

211

2

=

===LLL

LM


965

*77 •• Picture the Problem The figure shows the location of the cube’s center of mass and the forces acting on the cube. The opposing couple is formed by the friction force fs,max and the force exerted by the wall. Because the cube is in equilibrium, we can use the condition for translational equilibrium to establish that Wmaxs, Ff = and MgF =n

and the condition for rotational equilibrium to relate the opposing couples. Apply 0=∑F

r to the cube: MgFMgFFy =⇒=−=∑ nn 0

and

sWs 0 fFFfF Wx =⇒=−=∑

Noting that maxs,f

rand WF

rform a

couple, as do nFr

and ,gr

M apply 0=∑τr about an axis though the

center of mass of the cube:

0sinmaxs, =−Mgdaf θ

Referring to the diagram to the right, note

that ( )θ+°= 45sin2

ad .

Substitute for d and fs,max to obtain: ( ) 045sin

2sins =+°− θθµ aMgMga

or

( ) 045sin2

1sins =+°− θθµ

Solve for µs and simplify to obtain:

Chapter 12

966

( ) ( )

( )1cot21sin

21cos

21

sin21

sin45coscos45sinsin2145sin

sin21

s

+=⎟⎠⎞

⎜⎝⎛ +=

°+°=+°=

θθθθ

θθθ

θθ

µ

78 •• Picture the Problem Because the meter stick is in equilibrium, we can apply the condition for rotational equilibrium to find the maximum distance from the hinge at which the block can be suspended. Apply 0=∑τr about an axis through the hinge to obtain:

( )( ) ( )( )( ) ( )( ) 045cosm/s9.81kg0145cosm/s9.81kg5m0.5N75m1 22 =°−°− d

Solve for and evaluate d:

( )( ) ( )( )( )( )( ) m831.045cos

45cosm/s9.81kg01m/s9.81kg5m0.5N75m1

2

2

=°°

−=d

79 •• Picture the Problem Let m represent the mass of the ladder and M the mass of the person. The force diagram shows the forces acting on the ladder for part (b). From the condition for translational equilibrium, we can conclude that T = FW, a result we’ll need in part (b). Because the ladder is also in rotational equilibrium, summing the torques about the bottom of the ladder will eliminate both Fn and T.

(a) Apply 0=∑i iτ

r about an axis


( ) ( )( )( )( )( )( ) 0m/s9.81kg80m0.75

m/s9.81kg20m0.75m52

2W

=−

−F

Solve for and evaluate FW: ( )( )( )

( )( )( )

N147

m5m/s9.81kg80m0.75

m5m/s9.81kg20m0.75

2

2

W

=

+

=F


967

(b) Solve for and evaluate f :

( )( ) ( )( )( )( )( )( ) 724.0

m/s9.81kg80m5.1m/s9.81kg20m0.75N200m5

2

2

=−

=f

Find the distance the 80-kg person can climb the ladder:

( ) ( )( ) m62.3m5724.0m5 === fd

*80 •• Picture the Problem To ″roll″ the cube one must raise its center of mass from y = a/2 to 22ay = , where a is the cube

length. During this process the work done is the change in the gravitational potential energy of the cube. No additional work is done on the cube as it ″flops″ down. We can also use the definition of work to express the work done in sliding the cube a distance a along a horizontal surface and then equate the two expressions to determine µk.

Express the work done in moving the cube a distance a by raising its center of mass from y = a/2 to 22ay = and then letting the

cube flop down:

( )mga

mgaaamgW

207.0

12222

2

=

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

Letting fk represent the kinetic friction force, express the work done in dragging the cube a distance a along the surface at constant speed:

mgaafW kk µ==

Equate these two expressions to obtain: 207.0k =µ

Chapter 12

968

81 •• Picture the Problem The free-body diagram shows the forces acting on the block when it is on the verge of sliding. Because the block is in equilibrium, we can use the conditions for translational equilibrium to determine the minimum angle for which the block will slide. The diagram to the right of the FBD shows that the condition for tipping is that the plumb line from the center of mass pass outside of the base. We can determine the tipping angle from the geometry of the block under this condition.

Apply 0=∑F

r to the block: 0sin maxs,sliding ≥−=∑ fmgFx θ

if the block is to slide, and

∑ =−= 0cos slidingn θmgFFy

Substitute for fs,max and eliminate Fn between these equations to obtain:

slidings tanθµ ≤

Solve for the condition for sliding: ( ) ( ) °==≥ −− 8.214.0tantan 1s

1sliding µθ

Using the geometry of the block, express the condition on θ that must be satisfied if the block is to tip:

°=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛≥ −− 4.18

31tan

5.15.0tan 11

tipping aaθ

slides.it before tipblock will the, Because slidingtipping θθ <


969

82 •• Picture the Problem Let m represent the mass of the bar, M the mass of the suspended object, Fv the vertical component of the force the wall exerts on the bar, Fh the horizontal component of the force exerted the wall exerts on the bar, and T the tension in the cable. The free-body diagrams show these forces and their points of application on the bar for parts (a) and (b) of the problem. Because the bar is in equilibrium, we can apply the conditions for translational and rotational equilibrium to relate the various forces and distances.

(a) Apply 0=∑i iτ

r about an axis

through the hinge:

( ) ( )( ) 030cosm15

30cosm5.7m5=°−°−

MgmgT

Solve for T: ( ) ( )[ ]

m530cosm15m5.7 °+

=gMmT


( )( ) ( )( )

( )kN3.10

30cosm/s9.81m5

kg360m15kg85m5.7

2

=

°×

+=T

Apply 0=∑i iF

r to the bar: 060sinv =−−°+=∑ MgmgTFFy

and 060cosh =°−=∑ TFFx

Solve the y equation for Fv: ( )

( )( )( )kN55.4

m/s81.9kg360kg8560sinkN3.10

60sin

2

v

−=++

°−=++°−= gMmTF

Solve the x equation for Fh: ( )

kN15.560coskN3.1060cosh

=°=°= TF

Chapter 12

970

Find the magnitude of the force exerted by the wall on the bar:

( ) ( )kN6.87

kN5.15kN4.55 22

2h

2v

=

+−=

+= FFF

Find the direction of the force exerted by the wall on the bar:

°−=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

5.41

kN5.15kN4.55tantan 1

h

v1

FFθ

i.e., 41.5° below the horizontal.

(b) Apply 0=∑τr about the hinge: ( )[ ] ( )( ) 030cosm15

30cosm5.760sinm10=°−

°−°Mg

mgT

Solve for T: ( ) ( )

( ) °°

+= 30cos

60sinm10m15m5.7 gMmT


( )( ) ( )( )( )

( )kN92.5

30cosm/s81.960sinm10

kg360m15kg85m5.7

2

=

°×

°+

=T

Apply 0=∑F

r to the bar: ( )

( ) 0kg360

kg8560cosv

=−

−°+=∑g

gTFFy

and 060sinh =°−=∑ TFFx

Solve the y equation for Fv: ( )

( )( )kN41.1

m/s81.9kg360kg85

60coskN92.52

v

=++

°−=F

Solve the x equation for Fh: ( )

kN13.560sinkN92.560sinh

=°=°= TF

Find the magnitude of the force exerted by the wall on the bar:

( ) ( )kN32.5

kN5.13kN41.1 22

2h

2v

=

+=

+= FFF


971

Find the direction of the force exerted by the wall on the bar:

°=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

4.15

kN5.13kN1.41tantan 1

h

v1

FFθ

i.e., 15.4° above the horizontal. 83 •• Picture the Problem The box and the forces acting on it are shown in the figure. When the box is about to tip, Fn acts at its edge, as indicated in the drawing. We can use the definition of µs and apply the condition for rotational equilibrium in an accelerated frame to relate fs to the weight of the box and, hence, to the normal force.

Using its definition, express µs:

n

ss F

f≥µ


through the box’s center of mass:

0n21

s =− wFwf

Solve for the ratio n

s

Ff

: 21

n

s =Ff

Substitute to obtain the condition for tipping:

500.0s ≥µ

Therefore, if the box is to slide: 500.0s <µ , as in Problem 75.

Remarks: The difference between problems 75 and 83 is that in 75 the maximum acceleration before slipping is 0.5g, whereas in 88 it is (0.5 cos9°− sin9°) = 0.337g. *84 •• Picture the Problem Let the mass of the rod be represented by M. Because the rod is in equilibrium, we can apply the condition for rotational equilibrium to relate the masses of the objects placed on it to its mass.

Chapter 12

972


through the pivot for the initial condition:

( )( ) ( )( ) 0cm10

cm40g22cm20=−

−+M

mm

Solve for and evaluate M: ( )( ) ( )

g00.4

cm10cm40g22cm20

=

−+=

mmM


through the pivot for the second condition:

( ) ( ) 0cm10cm20 =− Mm

Solve for and evaluate m: ( ) g00.2cm20

cm1021 === MMm

*85 •• Picture the Problem Let the distance from the center of the meterstick of either finger be x1 and x2 and W the weight of the stick. Because the meterstick is in equilibrium, we can apply the condition for rotational equilibrium to obtain expressions for the forces one’s fingers exert on the meterstick as functions of the distances x1 and x2 and the weight of the meterstick W. We can then explain the stop-and-start motion of one’s fingers as they are brought closer together by considering the magnitudes of these forces in relationship the coefficients of static and kinetic friction.

(a)

finger.other at theoccurs slipping themeans which mass, ofcenter thecloser tofinger the

by exerted becan force frictional-staticlarger a ly,Consequent finger.other by the exertedan that greater th is mass ofcenter enearest th

finger by the exerted force normal stick the balanced aFor fingers. twoebetween th is mass ofcenter theas long as balanced remainsstick The


973

(b) Apply 0=∑τr about an axis through

point 1 to obtain:

( ) 01212 =−+ WxxxF

Solve for F2 to obtain: 21

12 xx

xWF+

=


through point 2 to obtain:

( ) 02211 =++− WxxxF

Solve for F1 to obtain: 21

21 xx

xWF+

=

not. isother theslipping isisfinger one When reversed. is process theand sliding, stops sliding that was

finger theslide, tobegins slidingnot t wasfinger tha point theAt that finger.other by the exerted force frictional-static maximum theexceeding force

frictional-kinetic a produce tolargely sufficient isstick on the exertsit force normal theuntil inward slide willmass ofcenter thefromfarthest finger The

86 •• Picture the Problem The drawing shows a side view of the wall-and-picture system. Because the frame’s width is not specified, we assume it to be negligible. Note that 0.75, 0.4, and 0.85 form a Pythagorean triad. Thus, the nail will be at the same level as the top of the frame. We can apply the condition for rotational equilibrium to determine the force exerted by the wall.

(a) Because the center of gravity of the picture is in front of the wall, the torque due to mg about the nail must be balanced by an opposing torque due to the force of the wall on the picture, acting horizontally. So that∑ = 0xF , the tension in the wire must have a

horizontal component, and the picture must therefore tilt forward. (b) Apply 0=∑τr about an axis

through the nail and parallel to the

( )[ ]( )( )( )[ ] 05cosm2.1

m/s81.9kg85sinm6.0

W

2

=°+°−

F

Chapter 12

974

wall to obtain:

Solve for and evaluate FW: ( )[ ]( )( )( )

N43.3

5cosm2.1m/s81.9kg85sinm6.0 2

W

=

°°

=F

87 •• Picture the Problem The box car and rail are shown in the drawing. At the critical speed, the normal force is entirely on the outside rail. The center of gravity is 0.775 m from that rail and 2.15 m above it. Choose the coordinate system shown in the figure. To find the speed at which this situation prevails, we can apply the conditions for static equilibrium in an accelerated frame.


through the center of gravity of the box car:

( ) ( ) 0m15.2m775.0 sn =− fF (1)

Apply ∑ = 0yF to the box car and

solve for Fn:

mgFmgF =⇒=− nn 0

Apply ∑ = cmmaFx to the box: Rvmf

2

s =

Substitute in equation (1) to obtain: ( ) ( ) 0m15.2m775.0

2

=−Rvmmg

Solve for v: Rgv 360.0=

(a) Evaluate v for R = 150 m: ( )( )

m/s0.23

m/s9.81m150360.0 2

=

=v


975

(b) Evaluate v for R = 240 m: ( )( )m/s1.29

m/s9.81m240360.0 2

=

=v

88 •• Picture the Problem For neutral equilibrium, the center of mass of the system must be at the same height as the feet of the tightrope walker. The system is shown in the drawing. Let the origin of the coordinate system be at the rope. We’ll determine the distance d such that ycm = 0. We’ll then determine the angle θ subtended by one half the long rod. Express the y coordinate of the center of mass of the system:

( )( ) ( )kg40kg58

kg202m0.9kg58cm +

−=

dy

Set ycm = 0 and solve for d:

d = 1.305 m

Relate the distances s and d and solve for s:

s = 0.65 m + d = 1.955 m

Relate s to R and θ: ( )θcos1−= Rs (1)

Relate R and θ to the half-length of the rod:

m4=θR (2)

Substitute in equation (1) to obtain: ( )θ

θcos1m4m955.1 −=

or

489.0cos1=

−θ

θ

Use graphical or trial-and-error methods to solve for θ :

rad08.1=θ

Substitute in equation (2) to obtain: m70.3rad1.08

m4==R

Chapter 12

976

*89 ••• Picture the Problem Let the mass of each brick be m and number them as shown in the diagrams for 3 bricks and 4 bricks below. Let l denote the maximum offset of the nth brick. Choose the coordinate system shown and apply the condition for rotational equilibrium about an axis parallel to the z axis and passing through the point P at the supporting edge of the nth brick.


through P and parallel to the z axis to bricks 1 and 2 for the 3-brick arrangement shown above on the left:

( )[ ] 021 =−+− ll mgLLmg

Solve for l to obtain:

L41=l


through P and parallel to the z axis to bricks 1 and 2 for the 4-brick arrangement shown above on the right:

( )[ ] ( )[ ]( ) 04

543

21

=−+−

+−++−

LLmgLLmgLLmg

l

ll

Solve for l to obtain:

L61=l

Continuing in this manner we obtain, as the successive offsets, the sequence:

nLLLLL2

...,8

,6

,4

,2

where n = 1, 2, 3, … N.

(c) Express the offset of the (n +1)st brick in terms of the offset of the nth brick:

nL

nn 21 +=+ ll


977

A spreadsheet program to calculate the sum of the offsets as a function of n is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Formula/Content Algebraic FormB5 B4+1 n + 1 C5 C4+$B$1/(2*B5)

nL

n 2+l

A B C D 1 L= 1 m 2 3 n offset 4 1 0.500 5 2 0.750 6 3 0.917 7 4 1.042 8 5 1.142 9 6 1.225

10 7 1.296 11 8 1.359 12 9 1.414 13 10 1.464

98 95 2.568 99 96 2.573

100 97 2.579 101 98 2.584 102 99 2.589 103 100 2.594

From the table we see that l5 = m,142.1 l10 = m,464.1 and

l100 = m.594.2

(d) Increasing N in the spreadsheet solution suggests that the sum of the individual offsets continues to grow as N increases without bound. The series is, in fact, divergent and the stack of bricks has no maximum offset or length.

Chapter 12

978

Offset as a function of n

0.0

0.5

1.0

1.5

2.0

2.5

3.0

0 20 40 60 80 100

n

Off

set

90 ••• Picture the Problem The four forces acting on the sphere: its weight, mg; the normal force of the plane, Fn; the frictional force, f, acting parallel to the plane; and the tension in the string, T, are shown in the figure. Choose the coordinate system shown. Because the sphere is in equilibrium, we can apply the conditions for translational and rotational equilibrium to find f, Fn, and T.


through the center of the sphere:

fTTRfR =⇒=− 0

Apply ∑ = 0xF to the sphere:

0sincos =−+ θθ MgTf

Substitute for f and solve for T: θθ

cos1sin

+=

MgT


( )( ) N89.7cos301

30sinm/s9.81kg3 2

=°+

°=T

(b) Apply ∑ = 0yF to the sphere: 0cossinn =−− θθ MgTF


979

Solve for Fn: θθ cossinn MgTF +=


( )( )( )

N4.29

30cosm/s9.81kg3

sin30N7.892

n

=

°+

°=F

(c) In part (a) we showed that f = T: N89.7=f

91 ••• Picture the Problem Let L be the length of each leg of the tripod. Applying the Pythagorean theorem leads us to conclude that the distance a shown in the figure is

L23

and the distance b, the distance to the

centroid of the triangle ABC is L23

32

, and

the distance c is3

L. These results allow

us to conclude that3

cos L=θ . Because

the tripod is in equilibrium, we can apply the condition for translational equilibrium to find the compressional forces in each leg.

Letting FC represent the compressional force in a leg of the tripod, apply

0=∑Fr

to the apex of the tripod:

0cos3 C =−mgF θ

Solve for FC: θcos3C

mgF =

Solve for FC: mgmgF

33

313

C =×

=

Substitute numerical values and evaluate FC: ( )( ) N566m/s9.81kg100

33 2

C ==F

Chapter 12

980

92 •• Picture the Problem The forces that act on the beam are its weight, mg; the force of the cylinder, Fc, acting along the radius of the cylinder; the normal force of the ground, Fn; and the friction force fs = µsFn. The forces acting on the cylinder are its weight, Mg; the force of the beam on the cylinder, Fcb = Fc in magnitude, acting radially inward; the normal force of the ground on the cylinder, Fnc; and the force of friction, fsc = µscFnc. Choose the coordinate system shown in the figure and apply the conditions for rotational and translational equilibrium.

Express µs,beam-floor in terms of fs and Fn:

n

sfloorbeams, F

f=−µ (1)

Express µs,cylinder-floor in terms of fsc and Fnc: nc

scfloorcylinders, F

f=−µ (2)


through the right end of the beam:

( )[ ] ( ) 0cm15coscm10 c =− Fmgθ

Solve for and evaluate Fc: ( )[ ]

[ ]( )( )

N3.2815

m/s9.81kg530cos10cm15coscm10

2

c

=

°=

=mgF θ

Apply ∑ = 0yF to the beam: ( ) 090coscn =−−°+ mgFF θ

Solve for Fn:

( )( ) ( )N5.42

cos30N28.3m/s9.81kg5

cos2

cn

=°−=

−= θFmgF

Apply ∑ = 0xF to the beam: ( ) 090coscs =−°+− θFf


981

Solve for and evaluate fs: ( ) ( )N2.14

cos60N28.390coscs

=°=−°= θFf

cbFr

is the reaction force to cFr

: N3.28ccb == FF radially inward.

Apply ∑ = 0yF to the cylinder: 0coscbnc =−− MgFF θ

Solve for and evaluate Fnc:

( ) ( )( )N103

m/s9.81kg8cos30N28.3

cos2

cbnc

=+°=

+= MgFF θ

Apply ∑ = 0xF to the cylinder: ( ) 090coscbsc =−°− θFf

Solve for and evaluate fsc: ( ) ( )

N14.2cos60N28.390coscbsc

=°=−°= θFf

Substitute numerical values in equations (1) and (2) and evaluate µs,beam-floor and µs,cylinder-floor:

580.0N24.5N14.2

floorbeams, ==−µ

and

138.0N031N14.2

floorcylinders, ==−µ

93 ••• Picture the Problem The geometry of the system is shown in the drawing. Let upward be the positive y direction and to the right be the positive x direction. Let the angle between the vertical center line and the line joining the two centers beθ. Then

rrR −

=θsin and( )RrR

rR−

−=

2tanθ .

The force exerted by the bottom of the cylinder is just 2mg. Let F be the force that the top sphere exerts on the lower sphere. Because the spheres are in equilibrium, we can apply the condition for translational equilibrium. Apply ∑ = 0yF to the spheres: 0n =−− mgmgF

Chapter 12

982

Solve for Fn: mgF 2n =

Because the cylinder wall is smooth, Fcosθ = mg, and: θcos

mgF =

Express the x component of F: θθ tansin mgFFx ==

Express the force that the wall of the cylinder exerts: ( )RrR

rRmgF−

−=

2W

Remarks: Note that as r approaches R/2, Fw→∞. *94 ••• Picture the Problem Consider a small rotational displacement, δθ of the cube from equilibrium. This shifts the point of contact between cube and cylinder by ,Rδθ where R = d/2. As a result of that motion, the cube itself is rotated through the same angle ,δθ and so its center is shifted in the same direction by the amount (a/2) ,δθ neglecting higher order terms in .δθ

If the displacement of the cube’s center of mass is less than that of the point of contact, the torque about the point of contact is a restoring torque, and the cube will return to its equilibrium position. If, on the other hand, (a/2)δθ > (d/2) ,δθ then the torque about the point of contact due to mg is in the direction of ,δθ and will cause the displacement from equilibrium to increase. We see that the minimum value of d/a for stable equilibrium is d/a = 1.

983

Chapter 13 Fluids Conceptual Problems 1 • Determine the Concept The absolute pressure is related to the gauge pressure according to P = Pgauge + Pat. While doubling the gauge pressure will increase the absolute pressure, we do not have enough information to say what the resulting absolute pressure will be. ( ) correct. is e

*2 •

Determine the Concept No. In an environment where 02

geff =−=rvmFg , there is no

buoyant force; there is no ″up″ or ″down.″ 3 •• Determine the Concept As you lower the rock into the water, the upward force you exert on the rock plus the upward buoyant force on the rock balance its weight. When the thread breaks, there will be an additional downward force on the scale equal to the buoyant force on the rock (the water exerts the upward buoyant force on the rock and the reaction force is the force the rock exerts on the water … and hence on the scale). Let ρ represent the density of the water, V the volume of the rock, and wf the weight of the displaced water. Then the density of the rock is 3ρ. We can use Archimedes’ principle to find the additional force on the scale. Apply Archimedes’ principle to the rock:

gVgmwB ffff ρ===

Because Vf = Vrock:

MggMgMB 31

rock 3===

ρρ

ρρ

and correct. is )(d

4 •• Determine the Concept The density of water increases with depth and the buoyant force on the rock equals the weight of the displaced water. Because the weight of the displaced water depends on the density of the water, it follows that the buoyant force on the rock increases as it sinks. correct. is )(b

5 •• Determine the Concept Nothing. The fish is in neutral buoyancy (that is, its density equals that water), so the upward acceleration of the fish is balanced by the downward

Chapter 13

984

acceleration of the displaced water. *6 •• Determine the Concept Yes. Because the volumes of the two objects are equal, the downward force on each side is reduced by the same amount when they are submerged, not in proportion to their masses. That is, if m1L1 = m2L2 and L1 ≠ L2, then (m1 − c)L1 ≠ (m2 − c)L2. 7 •• Determine the Concept The buoyant forces acting on these submerged objects are equal to the weight of the water each displaces. The weight of the displaced water, in turn, is directly proportional to the volume of the submerged object. Because ρPb > ρCu, the volume of the copper must be greater than that of the lead and, hence, the buoyant force on the copper is greater than that on the lead. correct. is )(b

8 •• Determine the Concept The buoyant forces acting on these submerged objects are equal to the weight of the water each displaces. The weight of the displaced water, in turn, is directly proportional to the volume of the submerged object. Because their volumes are the same, the buoyant forces on them must be the same. correct. is )(c

9 • Determine the Concept It blows over the ball, reducing the pressure above the ball to below atmospheric pressure. 10 • Determine the Concept From the equation of continuity (IV = Av = constant), we can conclude that, as the pipe narrows, the velocity of the fluid must increase. Using Bernoulli’s equation for constant elevation ( constant2

21 =+ vP ρ ), we can conclude

that as the velocity of the fluid increases, the pressure must decrease. correct. is )(b

*11 • Determine the Concept False. The buoyant force on a submerged object depends on the weight of the displaced fluid which, in turn, depends on the volume of the displaced fluid. 12 • Determine the Concept When the bottle is squeezed, the force is transmitted equally through the fluid, leading to a pressure increase on the air bubble in the diver. The air bubble shrinks, and the loss in buoyancy is enough to sink the diver.

Fluids

985

13 • Determine the Concept The buoyant force acting on the ice cubes equals the weight of the water they displace, i.e., .fff gVwB ρ== When the ice melts the volume of water

displaced by the ice cubes will occupy the space previously occupied by the submerged part of the ice cubes. Therefore, the water level remains constant. 14 • Determine the Concept The density of salt water is greater than that of fresh water and so the buoyant force exerted on one in salt water is greater than in fresh water. 15 •• Determine the Concept Because the pressure increases with depth, the object will be compressed and its density will increase. Its volume will decrease. Thus, it will sink to the bottom. 16 •• Determine the Concept The force acting on the fluid is the difference in pressure between the wide and narrow parts times the area of the narrow part. 17 •• Determine the Concept The drawing shows the beaker and a strip within the water. As is readily established by a simple demonstration, the surface of the water is not level while the beaker is accelerated, showing that there is a pressure gradient. That pressure gradient results in a net force on the small element shown in the figure.

*18 •• Determine the Concept The water level in the pond will drop slightly. When the anchor is in the boat, the boat displaces enough water so that the buoyant force on it equals the sum of the weight of the boat, your weight, and the weight of the anchor. When you put the anchor overboard, it will displace its volume and the volume of water displaced by the boat will decrease. 19 •• Determine the Concept From Bernoulli's principle, the opening above which the air flows faster will be at a lower pressure than the other one, which will cause a circulation of air in the tunnel from opening 1 toward opening 2. It has been shown that enough air will circulate inside the tunnel even with the slightest breeze outside.

Chapter 13

986

*20 • Determine the Concept The diagram that follows shows the forces exerted by the pressure of the liquid on the two cups to the left.

Because the force is normal to the surface of the cup, there is a larger downward component to the net force on the cup on the left. Similarly, there will be less total force exerted by the fluid in the cup on the far right in the diagram in the problem statement. Density 21 • Picture the Problem The mass of the cylinder is the product of its density and volume. The density of copper can be found in Figure 13-1. Using the definition of density, express the mass of the cylinder:

( )hRVm 2πρρ ==

Substitute numerical values and evaluate m:

( )( )( )

kg673.0

m106m102kg/m1093.8

2

2233

=

××

××=−

−πm

22 • Picture the Problem The mass of the sphere is the product of its density and volume. The density of lead can be found in Figure 13-1. Using the definition of density, express the mass of the sphere:

( )334 RVm πρρ ==


( )( )kg379.0

m102kg/m103.11 323334

=

××= −πm

Fluids

987

23 • Picture the Problem The mass of the air in the room is the product of its density and volume. The density of air can be found in Figure 13-1. Using the definition of density, express the mass of the air:

LWHVm ρρ ==


( )( )( )( )kg103

m4m5m4kg/m293.1 3

=

=m

*24 • Picture the Problem Let ρ0 represent the density of mercury at 0°C and ρ′ its density at 80°C, and let m represent the mass of our sample at 0°C and m′ its mass at 80°C. We can use the definition of density to relate its value at the higher temperature to its value at the lower temperature and the amount spilled. Using its definition, express the density of the mercury at 80°C: V

m'' =ρ

Express the mass of the mercury at 80°C in terms of its mass at 0°C and the amount spilled at the higher temperature: V

mVm

Vm

Vmm'

∆−=

∆−=

∆−=

0ρ

ρ

Substitute numerical values and evaluate ρ′:

34

36

334

kg/m101263.1

m1060kg101.47kg/m104563.1

×=

××

−×= −

−

'ρ

Pressure 25 • Picture the Problem The pressure due to a column of height h of a liquid of density ρ is given by P = ρgh. Letting h represent the height of the column of mercury, express the pressure at its base:

kPa101Hg =ghρ

Solve for h: g

hHg

kPa101ρ

=

Chapter 13

988

Substitute numerical values and evaluate h: ( )( )

Hg ofin 8.29

m102.54in1m0.757

m/s9.81kg/m1013.6N/m101.01

2

233

25

=

××=

××

=

−

h

26 • Picture the Problem The pressure due to a column of height h of a liquid of density ρ is given by P = ρgh. (a) Express the pressure as a function of depth in the lake:

ghPP waterat ρ+=

Solve for and evaluate h: g

PgPP

gPPh

water

at

water

atat

water

at 2ρρρ

=−

=−

=


m3.10

m/s81.9kg/m10N/m1001.1

233

25

=

×=h

(b) Proceed as in (a) with ρwater replaced by ρHg to obtain:

gP

gPPh

Hg

at

Hg

atat2ρρ

=−

=


cm7.57

m/s9.81kg/m1013.6N/m101.01

233

25

=

××

=h

*27 • Picture the Problem The pressure applied to an enclosed liquid is transmitted undiminished to every point in the fluid and to the walls of the container. Hence we can equate the pressure produced by the force applied to the piston to the pressure due to the weight of the automobile and solve for F. Express the pressure the weight of the automobile exerts on the shaft of the lift:

shaft

autoauto A

wP =

Express the pressure the force applied to the piston produces: pistonA

FP =

Fluids

989

Because the pressures are the same, we can equate them to obtain: pistonshaft

auto

AF

Aw

=

Solve for F:

shaft

pistonauto

shaft

pistonauto A

Agm

AA

wF ==

Substitute numerical values and evaluate F: ( )( )

N230

cm8cm1m/s9.81kg1500

22

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=F

28 •• Picture the Problem The pressure exerted by the woman’s heel on the floor is her weight divided by the area of her heel. Using its definition, express the pressure exerted on the floor by the woman’s heel:

Amg

Aw

AFP ===

Substitute numerical values and evaluate P:

( )( )

atm2.45

kPa3.101atm1N/m1049.5

m10m/s9.81kg56

26

24

2

=

××=

= −P

*29 • Picture the Problem The required pressure ∆P is related to the change in volume ∆V and

the initial volume V through the definition of the bulk modulus B; VV

PB∆∆

−= .

Using the definition of the bulk modulus, relate the change in volume to the initial volume and the required pressure:

VVPB

∆∆

−=

Solve for ∆P: VVBP ∆

−=∆

Chapter 13

990

Substitute numerical values and evaluate ∆P:

atm198

kPa101.325atm1Pa1000.2

L1L0.01Pa102.0

7

9

=

××=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −××−=∆P

30 • Picture the Problem The area of contact of each tire with the road is related to the weight on each tire and the pressure in the tire through the definition of pressure. Using the definition of gauge pressure, relate the area of contact to the pressure and the weight of the car:

gauge

41

PwA =

Substitute numerical values and evaluate A:

( )( )

( )( )

222

23

241

241

cm184m101.84

N/m10200m/s9.81kg1500

kPa200m/s9.81kg1500

=×=

×=

=

−

A

31 •• Picture the Problem The force on the lid is related to pressure exerted by the water and the cross-sectional area of the column of water through the definition of density. We can find the mass of the water from the product of its density and volume. (a) Using the definition of pressure, express the force exerted on the lid:

PAF =

Express the pressure due to a column of water of height h:

ghP waterρ=

Substitute for P and A to obtain:

2water rghF πρ=

Substitute numerical values: ( )( )( ) ( )

kN14.8

m0.2m12

m/s9.81kg/m102

233

=

×

=

π

F

Fluids

991

(b) Relate the mass of the water to its density and volume:

2waterwater rhVm πρρ ==


( )( ) ( )kg0.339

m103m12kg/m10 2333

=

×= −πm

32 •• Picture the Problem The minimum elevation of the bag h that will produce a pressure of at least 12 mmHg is related to this pressure and the density of the blood plasma through ghP bloodρ= .

Using the definition of the pressure due to a column of liquid, relate the pressure at its base to its height:

ghP bloodρ=

Solve for h: g

Phbloodρ

=


( )( )cm8.15m158.0

m/s9.81kg/m101.03mmHg1

Pa133.32mmHg12

233

==

×

×=h

33 •• Picture the Problem The depth h below the surface at which you would be able to breath is related to the pressure at that depth and the density of water ρw through ghP wρ= .

Express the pressure at a depth h and solve for h:

ghP wρ= and

gPhwρ

=

Express the pressure at depth h in terms of the weight on your chest: A

FP =


gAFh

wρ=

Chapter 13

992

Substitute numerical values and evaluate h: ( )( )( )

m0.453

m/s9.81kg/m10m0.09N400

2332

=

=h

34 •• Picture the Problem Let A1 and A2 represent the cross-sectional areas of the large piston and the small piston, and F1 and F2 the forces exerted by the large and on the small piston, respectively. The work done by the large piston is W1 = F1h1 and that done on the small piston is W2 = F2h2. We’ll use Pascal’s principle and the equality of the volume of the displaced liquid in both pistons to show that W1 and W2 are equal. Express the work done in lifting the car a distance h:

111 hFW =

where F is the weight of the car.

Using the definition of pressure, relate the forces F1 (= w) and F2 to the areas A1 and A2:

2

2

1

1

AF

AF

=

Solve for F1:

2

121 A

AFF =

Equate the volumes of the displaced fluid in the two pistons:

2211 AhAh =

Solve for h1:

1

221 A

Ahh =

Substitute in the expression for W1 and simplify to obtain: 222

1

22

2

121 WhF

AAh

AAFW ===

35 • Picture the Problem Because the pressure varies with depth, we cannot simply multiply the pressure times the half-area of a side of the cube to find the force exerted by the water. We therefore consider the force exerted on a strip of width a, height dh, and area dA = adh at a depth h and integrate from h = 0 to h = a/2. The water pressure at depth h is Pat + ρgh. We can

Fluids

993

omit the atmospheric pressure because it is exerted on both sides of the wall of the cube. Express the force dF on the element of length a and height dh in terms of the net pressure ρgh:

ghadhPdAdF ρ==

Integrate from h = 0 to h = a/2:

8

421

3

22

0

2

0

ga

agahdhgadFFaa

ρ

ρρ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=== ∫∫

*36 ••• Picture the Problem The weight of the water in the vessel is the product of its mass and the gravitational field. Its mass, in turn, is related to its volume through the definition of density. The force the water exerts on the base of the container can be determined from the product of the pressure it creates and the area of the base. (a) Using the definition of density, relate the weight of the water to the volume it occupies:

Vgmgw ρ==

Substitute for V to obtain: hgrw 231πρ=

Substitute numerical values and evaluate w:

( ) ( ) ( ) ( ) N8.57m/s9.81m1025m1015kg/m10 22223331 =××= −−πw

(b) Using the definition of pressure, relate the force exerted by the water on the base of the vessel to the pressure it exerts and the area of the base:

2rghPAF πρ==


( ) ( )( ) ( ) N173m1015m1025m/s9.81kg/m10 222233 =××= −− πF

Chapter 13

994

water.on the cone theof wallsslanting by the exerted force theofcomponent downward theofresult thealso

is base on the force downward The tube.in the water theof weight than thegreater much is barrel Pascalson force the way that same in the occurs This

Buoyancy *37 • Picture the Problem The scale’s reading will be the difference between the weight of the piece of copper in air and the buoyant force acting on it. Express the apparent weight w′ of the piece of copper:

Bww' −=

Using the definition of density and Archimedes’ principle, substitute for w and B to obtain:

( )VgVgVgw'

wCu

wCu

ρρρρ

−=−=

Express w in terms of ρCu and V and solve for Vg:

CuCu ρ

ρ wVgVgw =⇒=

Substitute to obtain: ( ) www' ⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

Cu

w

CuwCu 1

ρρ

ρρρ

Substitute numerical values and evaluate w′:

( )( )

N4.36

m/s9.81kg0.5911 2

=

⎟⎠⎞

⎜⎝⎛ −=w'

38 • Picture the Problem We can use the definition of density and Archimedes’ principle to find the density of the stone. The difference between the weight of the stone in air and in water is the buoyant force acting on the stone. Using its definition, express the density of the stone:

stone

stonestone V

m=ρ (1)

Apply Archimedes’ principle to obtain:

gVgmwB ffff ρ===

Fluids

995

Solve for Vf: g

BVf

f ρ=

Because Vf = Vstone and ρf = ρwater: g

BVwater

stone ρ=


waterstone

waterstone

stone ρρρB

wB

gm==

Substitute numerical values and evaluate ρstone:

( )33

33stone

kg/m1000.3

kg/m10N20N60

N60

×=

−=ρ

39 • Picture the Problem We can use the definition of density and Archimedes’ principle to find the density of the unknown object. The difference between the weight of the object in air and in water is the buoyant force acting on the object. (a) Using its definition, express the density of the object:

object

objectobject V

m=ρ (1)


gVgmwB ffff ρ===

Solve for Vf: g

BVf

f ρ=

Because Vf = Vobject and ρf = ρwater: g

BVwater

object ρ=


waterobject

waterobject

object ρρρB

wB

gm==

Substitute numerical values and evaluate ρobject:

( )33

33object

kg/m101.11

kg/m10N4.55N5

N5

×=

−=ρ

(b) lead. ofthat

toclosedensity a has materialunknown that thesee we1,-13 Figure From

Chapter 13

996

40 • Picture the Problem We can use the definition of density and Archimedes’ principle to find the density of the unknown object. The difference between the weight of the object in air and in water is the buoyant force acting on it. Using its definition, express the density of the metal:

metal

metalmetal V

m=ρ (1)


gVgmwB ffff ρ===

Solve for Vf: g

BVf

f ρ=

Because Vf = Vmetal and ρf = ρwater: g

BVwater

metal ρ=


watermetal

watermetal

metal ρρρB

wB

gm==

Substitute numerical values and evaluate ρmetal:

( )33

33metal

kg/m1069.2

kg/m10N56.6N90

0N9

×=

−=ρ

41 •• Picture the Problem Let V be the volume of the object and V′ be the volume that is submerged when it floats. The weight of the object is ρVg and the buoyant force due to the water is ρwV′g. Because the floating object is translational equilibrium, we can use

∑ = 0yF to relate the buoyant forces acting on the object in the two liquids to its

weight. Apply ∑ = 0yF to the object

floating in water:

0ww =−=− VgV'gmgV'g ρρρ (1)

Solve for ρ: VV'

wρρ =

Substitute numerical values and evaluate ρ:

( ) 333 kg/m8008.0kg/m10 ==V

Vρ

Fluids

997

Apply ∑ = 0yF to the object

floating in the second liquid and solve for mg:

gVmg L72.0 ρ=

Solve equation (1) for mg:

Vgmg w8.0 ρ=

Equate these two expressions to obtain:

wL 8.072.0 ρρ =

Substitute in the definition of specific gravity to obtain:

11.172.08.0gravityspecific

w

L ===ρρ

*42 •• Picture the Problem We can use Archimedes’ principle to find the density of the unknown object. The difference between the weight of the block in air and in the fluid is the buoyant force acting on the block. Apply Archimedes’ principle to obtain:

gVgmwB ffff ρ===

Solve for ρf: gV

Bf

f =ρ

Because Vf = VFe block: Fe

block Feblock Fef ρρ

gmB

gVB

==

Substitute numerical values and evaluate ρf:

( )( )( )( ) ( ) 3333

2

2

f kg/m1096.6kg/m1096.7m/s81.9kg5

N16.6m/s81.9kg5×=×

−=ρ

43 •• Picture the Problem The forces acting on the cork are B, the upward force due to the displacement of water, mg, the weight of the piece of cork, and Fs, the force exerted by the spring. The piece of cork is in equilibrium under the influence of these forces. Apply ∑ = 0yF to the piece of cork: 0s =−− FwB (1)

or 0scork =−− FVgB ρ (2)

Express the buoyant force as a function of the density of water:

VgwB wf ρ==

Chapter 13

998

Solve for Vg:

wρBVg =

Substitute for Vg in equation (2): 0s

wcork =−− FBB

ρρ (3)

Solve equation (1) for B: sFwB +=

Substitute in equation (3) to obtain: 0s

w

scorks =−

+−+ FFwFw

ρρ

or

0w

scork =

+−

ρρ Fww

Solve for ρcork:

swcork Fw

w+

= ρρ

Substitute numerical values and evaluate ρcork:

( )3

33cork

kg/m250

N0.855N0.285N0.285kg/m10

=

+=ρ

44 •• Picture the Problem Under minimum-volume conditions, the balloon will be in equilibrium. Let B represent the buoyant force acting on the balloon, wtot represent its total weight, and V its volume. The total weight is the sum of the weights of its basket, cargo, and helium in its balloon. Apply ∑ = 0yF to the balloon: 0tot =− wB

Express the total weight of the balloon: Vgw Hetot N2000 ρ+=

Express the buoyant force due to the displaced air:

VgwB airf ρ==

Substitute to obtain: 0N2000 Heair =−− VgVg ρρ

Solve for V:

( )gVHeair

N2000ρρ −

=

Fluids

999

Substitute numerical values and evaluate V:

( )( )3

233 m183m/s9.81kg/m0.178kg/m1.29

N2000=

−=V

*45 •• Picture the Problem Let V = volume of diver, ρD the density of the diver, VPb the volume of added lead, and mPb the mass of lead. The diver is in equilibrium under the influence of his weight, the weight of the lead, and the buoyant force of the water. Apply ∑ = 0yF to the diver: 0PbD =−− wwB

Substitute to obtain: 0PbDDPbDw =−−+ gmgVgV ρρ

or 0PbDDPbwDw =−−+ mVVV ρρρ

Rewrite this expression in terms of masses and densities:

0PbD

DD

Pb

Pbw

D

Dw =−−+ mmmm

ρρ

ρρ

ρρ

Solve for the mass of the lead: ( )

( )wPbD

DDwPbPb ρρρ

ρρρ−

−=

mm

Substitute numerical values and evaluate mPb:

( )( )( )( )( ) kg89.3

kg/m10kg/m1011.3kg/m100.96kg85kg/m100.96kg/m10kg/m1011.3

333333

333333

Pb =−××

×−×=m

46 •• Picture the Problem The scale’s reading w′ is the difference between the weight of the aluminum block in air w and the buoyant force acting on it. The buoyant force is equal to the weight of the displaced fluid, which, in turn, is the product of its density and mass. We can apply a condition for equilibrium to relate the reading of the bottom scale to the weight of the beaker and its contents and the buoyant force acting on the block. Express the apparent weight w′ of the aluminum block:

Bww' −= (1)

Letting F be the reading of the bottom scale and choosing upward to be the positive y direction, apply

0' tot =−+ gMwF (2)

Chapter 13

1000

0=∑ yF to the scale to obtain:

Using the definition of density and Archimedes’ principle, substitute for w and B in equation (1) to obtain:

( )VgVgVgw

wAl

wAl'ρρρρ

−=−=

Express w in terms of ρAl and V and solve for Vg:

AlAl ρ

ρ wVgVgw =⇒=

Substitute to obtain: ( ) www ⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

Al

w

AlwAl 1'

ρρ

ρρρ

Substitute numerical values and evaluate w′:

( )( )N4.12

m/s9.81kg2

kg/m102.7kg/m101'

2

33

33

=

×

⎟⎟⎠

⎞⎜⎜⎝

⎛×

−=w

Solve equation (2) for F:

'tot wgMF −=

Substitute numerical values and evaluate the reading of the bottom scale:

( )( )N7.36

N12.4m/s9.81kg5 2

=

−=F

47 ••• Picture the Problem Let V = displacement of ship in the two cases, m be the mass of ship without load, and ∆m be the load. The ship is in equilibrium under the influence of the buoyant force exerted by the water and its weight. We’ll apply the condition for floating in the two cases and solve the equations simultaneously to determine the loaded mass of the ship. Apply ∑ = 0yF to the ship in fresh water:

0w =−mgVgρ (1)

Apply ∑ = 0yF to the ship in salt water:

( ) 0sw =∆+− gmmVgρ (2)

Solve equation (1) for Vg:

wρmgVg =

Substitute in equation (2) to obtain: ( ) 0

wsw =∆+− gmmmgρ

ρ

Fluids

1001

Solve for m:

wsw

w

ρρρ−∆

=mm

Add ∆m to both sides of the equation and simplify to obtain:

wsw

sw

wsw

w

wsw

w

1

ρρρ

ρρρ

ρρρ

−∆

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−∆=

∆+−∆

=∆+

m

m

mmmm

Substitute numerical values and evaluate m + ∆m:

( )( )

( )( )

kg1046.21025.1

025.1kg106

025.1025.1kg106

7

5ww

w5

×=

−×

=

−×

=∆+ρρ

ρmm

*48 ••• Picture the Problem For minimum liquid density, the bulb and its stem will be submerged. For maximum liquid density, only the bulb is submerged. In both cases the hydrometer will be in equilibrium under the influence of its weight and the buoyant force exerted by the liquids. (a) Apply ∑ = 0yF to the hydrometer: 0=− wB

Using Archimedes’ principle to express B, substitute to obtain:

0totmin =− gmVgρ

or ( ) Pbglassstembulbmin mmVV +=+ρ

Solve for mPb: ( ) glassstembulbminPb mVVm −+= ρ


( ) ( )( ) ( )

g7.14

kg106m10

L1m005.0m15.04

L020.0kg/L9.0 333-

2Pb

=

×−⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+= −πm

(b) Apply ∑ = 0yF to the hydrometer: 0totmax =− gmVgρ

or

Chapter 13

1002

Pbglassbulbmax mmV +=ρ

Solve for ρmax:

bulb

Pbglassmax V

mm +=ρ

Substitute numerical values and evaluate ρmax:

kg/L04.1mL20

g14.7g6max =

+=ρ

49 • Picture the Problem We can relate the upward force exerted on the dam wall to the area over which it acts using APF g= and express Pg in terms of the depth of the water

using ghP ρ=g . Using the definition of pressure, express the upward force exerted on the dam wall:

APF g=

Express the gauge pressure Pg of the water 5 m below the surface of the dam:

ghP ρ=g


ghAF ρ=


( )( )( )( )kN491

m10m5m/s81.9kg/m10 2233

=

=F

50 •• Picture the Problem The forces acting on the balloon are the buoyant force B, its weight mg, and a drag force FD. We can find the initial upward acceleration of the balloon by applying Newton’s 2nd law at the instant it is released. We can find the terminal velocity of the balloon by recognizing that when ay = 0, the net force acting on the balloon will be zero. (a) Apply ∑ = yy maF to the balloon at the instant of its release to obtain:

yamgmB balloonballoon =−

Solve for ay: gm

Bm

gmBay −=−

=balloonballoon

balloon

Fluids

1003

Using Archimedes principle, express the buoyant force B acting on the balloon:

grgV

gVgmwB3

air34

balloonair

ffff

πρρ

ρ

==

===


yamgmgr balloonballoon3

air34 =−πρ

Solve for ay:

gm

ray ⎟⎟

⎠

⎞⎜⎜⎝

⎛−= 1

balloon

3air3

4 πρ

Substitute numerical values and evaluate ay:

( )( )

( )2

2

3334

m/s4.45

m/s81.9

1kg15

m5.2kg/m29.1

=

×

⎥⎦

⎤⎢⎣

⎡−=

πya

(b) Apply ∑ = yy maF to the balloon under terminal-speed conditions to obtain:

02t

221 =−− vrmgB ρπ

Substitute for B:

02t

2213

air34 =−− vrmggr ρππρ

Solve for vt: ( )ρπ

πρ2

3air3

4

t

2r

gmrv

−=


( )( )[ ]( )( ) ( ) m/s33.7

kg/m29.1m5.2m/s81.9kg15m5.2kg/m29.12

32

23334

t =−

=π

πv

(c) Relate the time required for the balloon to rise to 10 km to its terminal speed: min7.22

s1364m/s33.7

km10

=

===∆tv

ht

Continuity and Bernoulli's Equation *51 •• Picture the Problem Let J represent the flow rate of the water. Then we can use J = Av to relate the flow rate to the cross-sectional area of the circular tap and the velocity of the water. In (b) we can use the equation of continuity to express the diameter of the stream 7.5 cm below the tap and a constant-acceleration equation to find the velocity of the water at this distance. In (c) we can use a constant-acceleration equation to express the distance-to-turbulence in terms of the velocity of the water at turbulence vt and the definition of Reynolds number NR to relate vt to NR.

Chapter 13

1004

(a) Express the flow rate of the water in terms of the cross-sectional area A of the circular tap and the velocity v of the water:

vdvrAvJ 2412 ππ === (1)

Solve for v: 2

41 d

Jvπ

=

Substitute numerical values and evaluate v: ( )

cm/s28.9cm2.1

s/cm5.102

41

3

==π

v

(b) Apply the equation of continuity to the stream of water:

iiiff vAAvAv == or

2i

2ff 44

dvdv ππ=

Solve for df: i

ff d

vvd = (2)

Use a constant-acceleration equation to relate vf and v to the distance ∆h fallen by the water:

hgvv ∆+= 222f

Solve for vf to obtain:

hgvv ∆+= 22f

Substitute numerical values and evaluate vf:

( ) ( )( )cm/s122

cm5.7cm/s9812cm/s28.9 22f

=

+=v

Substitute in equation (2) and evaluate df: ( ) cm331.0

cm/s122cm/s28.9cm2.1f ==d

(c) Using a constant-acceleration equation, relate the fall-distance-to-turbulence ∆d to its initial speed v and its speed vt when its flow becomes turbulent:

dgvv ∆+= 222t

Solve for ∆d to obtain: g

vvd2

22t −=∆ (3)

Express Reynolds number NR for turbulent flow: η

ρ tR

2 vrN =

Fluids

1005

From equation (1):

tvJrπ

=


t

tR

2v

JvNπη

ρ=

Solve for vt:

JNv 2

22R

t 4ρηπ

=

Substitute numerical values (see Figure 13-1 for the density of water and Table 13-1 for the coefficient of viscosity for water) and evaluate vt:

( ) ( )( ) ( )

m/s28.1s/cm5.10kg/m104

sPa108.123003233

232

t

=

⋅×=

−πv

Substitute in equation (3) and evaluate the fall-distance-to turbulence:

( ) ( )( )

cm31.8

cm/s9812cm/s28.9cm/s128

2

22

=

−=∆d

in reasonable agreement with everyday experience.

52 • Picture the Problem Let A1 represent the cross-sectional area of the hose, A2 the cross-sectional area of the nozzle, v1 the velocity of the water in the hose, and v2 the velocity of the water as it passes through the nozzle. We can use the continuity equation to find v2 and Bernoulli’s equation for constant elevation to find the pressure at the pump. (a) Using the continuity equation, relate the speeds of the water to the diameter of the hose and the diameter of the nozzle:

2211 vAvA =

or

2

22

1

21

44vdvd ππ

=

Solve for v2:

122

21

2 vddv =

Substitute numerical values and evaluate v2: ( ) m/s0.65m/s0.65

cm0.3cm3

2

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛=v

(b) Using Bernoulli’s equation for constant elevation, relate the pressure at the pump PP to the

222

1at

212

1P vPvP ρρ +=+

Chapter 13

1006

atmospheric pressure and the velocities of the water in the hose and the nozzle:

Solve for the pressure at the pump: ( )21

222

1atP vvPP −+= ρ

Substitute numerical values and evaluate PP:

( ) ( ) ( )[ ]atm21.9

kPa101.325atm1Pa1021.2

m/s0.65m/s65kg/m10kPa101

6

223321

P

=××=

−+=P

53 • Picture the Problem Let A1 represent the cross-sectional area of the larger-diameter pipe, A2 the cross-sectional area of the smaller-diameter pipe, v1 the velocity of the water in the larger-diameter pipe, and v2 the velocity of the water in the smaller-diameter pipe. We can use the continuity equation to find v2 and Bernoulli’s equation for constant elevation to find the pressure in the smaller-diameter pipe. (a) Using the continuity equation, relate the velocities of the water to the diameters of the pipe:

2211 vAvA =

or

2

22

1

21

44vdvd ππ

=

Solve for and evaluate v2:

122

21

2 vddv =

Substitute numerical values and evaluate v2: ( ) m/s0.12m/s3

2

121

12 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=

ddv

(b) Using Bernoulli’s equation for constant elevation, relate the pressures in the two segments of the pipe to the velocities of the water in these segments:

22w2

12

21w2

11 vPvP ρρ +=+

Solve for P2: ( )2

221w2

11

22w2

121w2

112

vvP

vvPP

−+=

−+=

ρ

ρρ

Fluids

1007

Substitute numerical values and evaluate P2:

( )( ) ( )[ ]

kPa133

m/s12m/s3

kg/m10kPa20022

3321

2

=

−×

+=P

(c) Using the continuity equation, evaluate IV1:

( )m/s344

21

1

21

11V1dvdvAI ππ

===

Using the continuity equation, express IV2: 2

22

22V2 4vdvAI π

==

Substitute numerical values and evaluate IV2: ( ) ( )m/s3

4m/s12

42 2

1

21

V2d

d

I ππ

=⎟⎠⎞

⎜⎝⎛

=

Thus, as we expected would be the case:

V2V1 II =

54 • Picture the Problem Let A1 represent the cross-sectional area of the 2-cm diameter pipe, A2 the cross-sectional area of the constricted pipe, v1 the velocity of the water in the 2-cm diameter pipe, and v2 the velocity of the water in the constricted pipe. We can use the continuity equation to express d2 in terms of d1 and to find v1 and Bernoulli’s equation for constant elevation to find the velocity of the water in the constricted pipe. Using the continuity equation, relate the volume flow rate in the 2-cm diameter pipe to the volume flow rate in the constricted pipe:

2211 vAvA =

or

2

22

1

21

44vdvd ππ

=

Solve for d2:

2

112 v

vdd =

Using the continuity equation, relate v1 to the volume flow rate IV: ( )

m/s91.8

4m0.02

L/s80.22

1

V1 ===

πAIv

Using Bernoulli’s equation for constant elevation, relate the pressures in the two segments of the pipe to the velocities of the water in

22w2

12

21w2

11 vPvP ρρ +=+

Chapter 13

1008

these segments:

Solve for v2: ( ) 21

w

12

2 vPPv +−

=ρ


( ) ( )

m/s7.12

m/s8.91kg/m10

kPa101kPa1422 2332

=

+−

=v

Substitute and evaluate d2: ( ) cm68.1

m/s12.7m/s8.91cm22 ==d

*55 •• Picture the Problem We can use the definition of the volume flow rate to find the volume flow rate of blood in an aorta and to find the total cross-sectional area of the capillaries. (a) Use the definition of the volume flow rate to find the volume flow rate through an aorta:

AvI =V

Substitute numerical values and evaluate IV:

( )( )

L/min58.4

m10L1

mins60

sm1063.7

m/s3.0m109

33

35

33V

=

×××=

×=

−−

−πI

(b) Use the definition of the volume flow rate to express the volume flow rate through the capillaries:

capcapV vAI =

Solve for the total cross-sectional area of the capillaries: cap

Vcap v

IA =

Substitute numerical values and evaluate Acap:

222

35

cap

cm763m107.63

m/s001.0/sm107.63

=×=

×=

−

−

A

Fluids

1009

56 •• Picture the Problem We can apply Bernoulli’s equation to points a and b to determine the rate at which the water exits the tank. Because the diameter of the small pipe is much smaller than the diameter of the tank, we can neglect the velocity of the water at the point a. The distance the water travels once it exits the pipe is the product of its speed and the time required to fall the distance H – h. Express the distance x as a function of the exit speed of the water and the time to fall the distance H – h:

tvx b∆= (1)

Apply Bernoulli’s equation to the water at points a and b:

( ) 2w2

1w

b2

w21

w

b

aa

vhHg

PvgHP

ρρ

ρρ

+−+

=++

or, because va ≈ 0 and Pa = Pb = Pat, ( ) 2

21

bvhHggH +−=

Solve for vb: ghvb 2=

Using a constant-acceleration equation, relate the time of fall to the distance of fall:

( )221

0y tatvy ∆+∆=∆

or, because v0y = 0, ( )22

1 tghH ∆=−

Solve for ∆t: ( )

ghHt −

=∆2

Substitute in equation (1) to obtain: ( ) ( )hHh

ghHghx −=

−= 222

57 •• Picture the Problem Let the subscript 60 denote the 60-cm-radius pipe and the subscript 30 denote the 30-cm-radius pipe. We can use Bernoulli’s equation for constant elevation to express P30 in terms of v60 and v30, the definition of volume flow rate to find v60 and the continuity equation to find v30. Using Bernoulli’s equation for constant elevation, relate the pressures in the two pipes to the velocities of the oil:

2302

130

2602

160 vPvP ρρ +=+

Chapter 13

1010

Solve for P30: ( )230

2602

16030 vvPP −+= ρ (1)

Use the definition of volume flow rate to find v60:

( )m/s456.2

m6.0s3600

h1h24

day1daym102.4

2

35

60

V60

=

×××=

=

π

AIv

Using the continuity equation, relate the velocity of the oil in the half-standard pipe to its velocity in the standard pipe:

30306060 vAvA =

Solve for and evaluate v30: ( )( )

( )

m/s824.9

m/s456.2m3.0m6.0

2

2

6030

6030

=

==ππv

AAv

Substitute numerical values in equation (1) and evaluate P30:

( ) ( ) ( )[ ] kPa144m/s824.9m/s456.2kg/m800kPa180 22321

30 =−+=P

*58 •• Picture the Problem We’ll use its definition to relate the volume flow rate in the pipe to the velocity of the water and the result of Example 13-9 to find the velocity of the water. Using its definition, express the volume flow rate:

12

11V vrvAI π==

Using the result of Example 13-9, find the velocity of the water upstream from the Venturi meter: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=1

2

22

21

w

Hg1

RR

ghv

ρ

ρ


( )( )( )

( )m/s847.1

1m0.056m0.095kg/m10

m0.024m/s9.81kg/m1013.622

33

233

1 =

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛

×=v

Fluids

1011

Substitute numerical values and evaluate IV:

( ) ( )

L/s1.13

/sm10309.1

m/s1.847m095.04

32

2V

=

×=

=

−

πI

59 •• Picture the Problem We can apply the definition of the volume flow rate to find the mass of water emerging from the hose in 1 s and the definition of momentum to find the momentum of the water. The force exerted on the water by the hose can be found from the rate at which the momentum of the water changes. (a) Using its definition, express the volume flow rate of the water emerging from the hose:

Avt

mtVI =

∆∆

=∆∆

=w

V ρ

Solve for ∆m: tAvm ∆=∆ wρ

Substitute numerical values and evaluate ∆m:

( ) ( )( )( )kg/s2.21

s1kg/m10m/s30m015.0 332

=

=∆ πm

(b) Using its definition, express and evaluate the momentum of the water:

( )( )m/skg636

m/s30kg/s21.2

⋅=

=∆= mvp

(c) The vector diagrams are to the right:

Express the change in momentum of the water:

if ppp rrr−=∆

Substitute numerical values and evaluate ∆p:

( ) ( )( )

m/skg899

2m/skg636

m/skg636m/skg636 22

⋅=

⋅=

⋅+⋅=∆p

Chapter 13

1012

Relate the force exerted on the water by the hose to the rate at which the water’s momentum changes and evaluate F:

N899s1

m/skg998=

⋅=

∆∆

=tpF

60 •• Picture the Problem Let the letter P denote the pump and the 2-cm diameter pipe and the letter N the 1-cm diameter nozzle. We’ll use Bernoulli’s equation to express the necessary pump pressure, the continuity equation to relate the velocity of the water coming out of the pump to its velocity at the nozzle, and a constant-acceleration equation to relate its velocity at the nozzle to the height to which the water rises. Using Bernoulli’s equation, relate the pressures, areas, and velocities in the pipe and nozzle:

2Nw2

1

NwN2Pw2

1PwP

v

ghPvghP

ρ

ρρρ

+

+=++

or, because PN = Pat and hP = 0, 2Nw2

1NwN

2Pw2

1P vghPvP ρρρ ++=+

Solve for the pump pressure: ( )2

P2Nw2

1NwatP vvghPP −++= ρρ (1)

Use the continuity equation to relate vP and vN to the cross-sectional areas of the pipe from the pump and the nozzle:

NNPP vAvA =

and

N41

N

2

N2P4

1

2N4

1

NP

NP cm2

cm1

v

vvddv

AAv

=

⎟⎟⎠

⎞⎜⎜⎝

⎛===

ππ

Using a constant-acceleration equation, express the velocity of the water at the nozzle in terms of the desired height ∆h:

hgvv ∆−= 222N

or, because v = 0, hgv ∆= 22

N


( )[ ] ( )( )hhgP

hgghPhghgghPP∆++=

∆++=∆−∆++=

1615

Nwat

815

w21

Nwat161

w21

NwatP 22ρ

ρρρρ

Substitute numerical values and evaluate PP:

( )( ) ( )[ ] kPa241m12m3m/s9.81kg/m10kPa101 1615233

P =++=P

Fluids

1013

61 ••• Picture the Problem We can apply Bernoulli’s equation to points a and b to determine the rate at which the water exits the tank. Because the diameter of the small pipe is much smaller than the diameter of the tank, we can neglect the velocity of the water at the point a. The distance the water travels once it exits the pipe is the product of its velocity and the time required to fall the distance H – h. That there are two values of h that are equidistant from the point Hh 2

1= can be shown by solving the quadratic equation that

relates x to h and H. That x is a maximum for this value of h can be established by treating x = f(h) as an extreme-value problem. (a) Express the distance x as a function of the exit speed of the water and the time to fall the distance H – h:

tvx b∆= (1)

Apply Bernoulli’s equation to the water at points a and b:

( )2

w21

wb2

w21

w

b

aa

v

hHgPvgHP

ρ

ρρρ

+

−+=++

or, because va ≈ 0 and Pa = Pb = Pat, ( ) 2

21

bvhHggH +−=

Solve for vb: ghvb 2=

Using a constant-acceleration equation, relate the time of fall to the distance of fall:

( )221

0y tatvy ∆+∆=∆

or, because v0y = 0, ( )22

1 tghH ∆=−

Solve for ∆t: ( )

ghHt −

=∆2

Substitute in equation (1) to obtain: ( ) ( )hHh

ghHghx −=

−= 222

(b) Square both sides of this equation and simplify to obtain:

22 44 hhHx −= or 044 22 =+− xHhh

Solve this quadratic equation to obtain:

2221

21 xHHh −±=

Chapter 13

1014

Find the average of these two values for h: H

xHHxHHh

21

222122

21

av 2=

−−+−+=

(c) Differentiate

( )hHhx −= 2 with respect to h: ( )[ ] ( )

( )hHhhH

hHhHhdhdx

−

−=

−−⎟⎠⎞

⎜⎝⎛= −

2

2212 2

1

Set the derivative equal to zero for extrema:

( )02

=−

−hHh

hH

Solve for h to obtain:

Hh 21=

Evaluate ( )hHhx −= 2 with Hh 2

1= : ( )

H

HHHx

=

−= 21

21

max 2

Remarks: To show that this value for h corresponds to a maximum, one can either

show that 0<2

2

dhxd

at Hh 21= or confirm that the graph of f(h) at Hh 2

1= is

concave downward. *62 •• Picture the Problem Let the numeral 1 denote the opening in the end of the inner pipe and the numeral 2 to one of the holes in the outer tube. We can apply Bernoulli’s principle at these locations and solve for the pressure difference between them. By equating this pressure difference to the pressure difference due to the height h of the liquid column we can express v as a function of ρ, ρg, g, and h.

Apply Bernoulli’s principle at locations 1 and 2 to obtain:

22g2

12

21g2

11 vPvP ρρ +=+

where we’ve ignored the difference in elevation between the two openings.

Solve for the pressure difference ∆P = P1 − P2:

21g2

122g2

121 vvPPP ρρ −=−=∆

Express the velocity of the gas at 1: 01 =v because the gas is brought to a halt (i.e., is stagnant) at the opening to the inner pipe.

Express the velocity of the gas at 2: vv =2 because the gas flows freely past

Fluids

1015

the holes in the outer ring.

Substitute to obtain: 2g2

1 vP ρ=∆

Letting A be the cross-sectional area of the tube, express the pressure at a depth h in the column of liquid whose density is ρ1:

AB

Aw

PP −+= liquid displaced21

where AhgB gρ= is the buoyant force acting on the column of liquid of height h.


( )ghPAghA

AghAPP

g2

g21

ρρ

ρρ

−+=

−+=

or ( )ghPPP g21 ρρ −=−=∆

Equate these two expressions for ∆P:

( )ghv g2

g21 ρρρ −=


( )g

g2 2ρ

ρρ −=

ghv

Note that the correction for buoyant force due to the displaced gas is very small and that, to a good approximation,

.2gρρghv =

Remarks: Pitot tubes are used to measure the airspeed of airplanes. 63 •• Picture the Problem Let the letter ″a″ denote the entrance to the siphon tube and the letter ″b″ denote its exit. Assuming streamline flow between these points, we can apply Bernoulli’s equation to relate the entrance and exit speeds of the water flowing in the siphon to the pressures at either end, the density of the water, and the difference in elevation between the entrance and exit points. We can use the expression for the pressure as a function of depth in an incompressible fluid to find the pressure at the entrance to the tube in terms of its distance below the surface. We’ll also use the equation of continuity to argue that, provided the surface area of the beaker is large compared to the area of the opening of the tube, the entrance speed of the water is approximately zero.

(a) Apply Bernoulli’s equation at the entrance to the siphon tube (point a) and at its exit (point b):

( )( )dhHgvP

hHgvP

b −−++=

−++

ρρ

ρρ2

21

b

2a2

1a (1)

where H is the height of the containers.

Chapter 13

1016

Apply the continuity equation to a point at the surface of the liquid in the container to the left and to point a:

surfacesurfaceaa AvAv = or, because Aa << Asurface,

0surfacea == vv

Express the pressure at the inlet (point a) and the outlet (point b):

( )hHgPP −+= ρatma and

( )dhHgPP −−+= ρatmb

Letting vb = v, substitute in equation (1) to obtain:

( )( )

( )dhHgv

dhHgPgHhHgP

−−++

−−+=+−+

ρρ

ρρρ

221

atm

atm

or, upon simplification, ( ) ( )

( )dhHgvdhHggHhHg

−−++

−−=+−2

21

Solve for v: gdv 2=

(b) Relate the pressure at the highest part of the tube Ptop to the pressure at point b:

( )( ) 2

b21

atm

2h2

1top

vdhHgP

vhHgP

ρρ

ρρ

+−−+=

+−+

or, because vh = vb, gdPP ρ−= atmtop

Remarks: If we let Ptop = 0, we can use this result to find the maximum theoretical height a siphon can lift water. Viscous Flow 64 • Picture the Problem The required pressure difference can be found by applying Poiseuille’s law to the viscous flow of water through the horizontal tube. Using Poiseuille’s law, relate the pressure difference between the two ends of the tube to its length, radius, and the volume flow rate of the water:

V4

8 IrLP

πη

=∆


( )( )( ) ( )

kPa47.1

mL/s3.0m106.0

m25.0smPa1843

=

×

⋅=∆

−πP

Fluids

1017

65 • Picture the Problem Because the pressure difference is unchanged, we can equate the expressions of Poiseuille’s law for the two tubes and solve for the diameter of the tube that would double the flow rate. Using Poiseuille’s law, express the pressure difference required for the radius and volume flow rate of Problem 64:

V4

8 IrLP

πη

=∆

Express the pressure difference required for the radius r′ that would double the volume flow rate of Problem 57:

( )V4 28 Ir'LP

πη

=∆

Equate these equations and simplify to obtain:

( ) V4V4828 I

rLI

r'L

πη

πη

=

or

4412rr'

=

Solve for r′: rr' 4 2=

Express d ′:

drr'd' 44 2222 ===

Substitute numerical values and evaluate d ′:

( ) mm43.1mm2.124 ==d'

*66 • Picture the Problem We can apply Poiseuille’s law to relate the pressure drop across the capillary tube to the radius and length of the tube, the rate at which blood is flowing through it, and the viscosity of blood. Using Poiseuille’s law, relate the pressure drop to the length and diameter of the capillary tube, the volume flow rate of the blood, and the viscosity of the blood:

V4

8 IrLP

πη

=∆

Solve for the viscosity of the blood:

V

4

8LIPr ∆

=πη

Chapter 13

1018

Using its definition, express the volume flow rate of the blood:

vrvAI 2capV π==

Substitute and simplify:

LvPr

8

2∆=η

Substitute numerical values to obtain: ( ) ( )

( )

smPa98.3

s1m10m108

kPa60.2m105.33

3

26

⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

=−

−

−

η

*67 • Picture the Problem We can use the definition of Reynolds number to find the velocity of a baseball at which the drag crisis occurs. Using its definition, relate Reynolds number to the velocity v of the baseball:

ηρvrN 2

R =

Solve for v:

ρη

rNv

2R=

Substitute numerical values (see Figure 13-1 for the density of air and Table 13-1 for the coefficient of viscosity for air) and evaluate v:

( )( )( )( )

mi/h93.4

m/s0.447mi/h1m/s8.41

kg/m293.1m05.02103smPa018.0

3

5

=

×=

×⋅=v

game. in the role aplay ellmay very w crisis drag this 90s,-mid to-low in the fastball a can throw pitchers leaguemajor most Because

Remarks: This is a topic which has been fiercely debated by people who study the physics of baseball.

Fluids

1019

68 ••• Picture the Problem Let the subscripts ″f″ refer to ″displaced fluid”, ″s″ to ″soda″, and ″g″ to the ″gas″ in the bubble. The free-body diagram shows the forces acting on the bubble prior to reaching its terminal velocity. We can apply Newton’s 2nd law, Stokes′ law, and Archimedes principle to express the terminal velocity of the bubble in terms of its radius, and the viscosity and density of water.

Apply yy maF =∑ to the bubble to obtain:

ymaFgmB =−− Dg

Under terminal speed conditions: 0Dg =−− FgmB

Using Archimedes principle, express the buoyant force B acting on the bubble:

gVgVgmwB

bubblesff

ff

ρρ ====

Express the mass of the gas bubble:

bubblegggg VVm ρρ ==

Substitute to obtain: 06 tbubblegbubblew =−− avgVgV πηρρ

Solve for vt: ( )a

gVv

πηρρ

6gsbubble

t

−=

Substitute for Vbubble and simplify: ( ) ( )

. since ,9

29

26

gss

2

gs2

gs3

34

t

ρρηρ

ηρρ

πηρρπ

>>≈

−=

−=

ga

gaa

gav

Substitute numerical values and evaluate vt:

( ) ( )( )

( )m/s333.0

kg/m101.1sPa108.19m/s81.9m105.02

33

3

223

t

=

××

⋅××

= −

−

v

Express the rise time ∆t in terms of the height of the soda glass h and the terminal speed of the bubble:

tvht =∆

Chapter 13

1020

Assuming that a "typical" soda glass has a height of about 15 cm, evaluate ∆t:

s450.0m/s333.0m15.0

==∆t

Remarks: About half a second seems reasonable for the rise time of the bubble. General Problems *69 •• Picture the Problem We can solve the given equation for the coefficient of roundness C and substitute estimates/assumptions of typical masses and heights for adult males and females. Express the mass of a person as a function of C, ρ, and h:

3hCM ρ=

Solve for C: 3h

MCρ

=

Assuming that a "typical" adult male stands 5' 10" (1.78 m) and weighs 170 lbs (77 kg), then:

( )( )0137.0

m78.1kg/m10kg77

333 ==C

Assuming that a "typical" adult female stands 5' 4" (1.63 m) and weighs 110 lbs (50 kg), then: ( )( )

0115.0m63.1kg/m10

kg50333 ==C

70 • Picture the Problem Let the letter ″s″ denote the shorter of the two men and the letter ″t″ the taller man. We can find the difference in weight of the two men using the relationship M = Cρh3 from Problem 69. Express the difference in weight of the two men: ( )gMM

gMgMwww

st

stst

−=−=−=∆

Express the masses of the two men:

3ss hCM ρ=

and 3tt hCM ρ=


( ) gChh

ghChCw

ρ

ρρ3s

3t

3s

3t

−=

−=∆

Assuming that a "typical" adult male stands 5' 10" (1.78 m) and weighs 170 lbs (77 kg), then:

( )( )0137.0

m78.1kg/m10kg77

333 ==C

Express the heights of the two men in SI units:

m1.83cm/in54.2in72t =×=h and

Fluids

1021

m1.75cm/in54.2in69s =×=h

Substitute numerical values (assume that ρ = 103 kg/m3) and evaluate ∆w:

( ) ( )[ ]( )( )( )

lb2.23N4.4482

lb1N103

m/s81.9kg/m100137.0m75.1m83.1

233

33

=×=

×

−=∆w

71 • Determine the Concept The net force is zero. Neglecting the thickness of the table, the atmospheric pressure is the same above and below the surface of the table. 72 • Picture the Problem The forces acting on the Ping-Pong ball, shown in the free-body diagram, are the buoyant force, the weight of the ball, and the tension in the string. Because the ball is in equilibrium under the influence of these forces, we can apply the condition for translational equilibrium to establish the relationship between them. We can also apply Archimedes’ principle to relate the buoyant force on the ball to its diameter.

Apply ∑ = 0yF to the ball: 0=−− TmgB

Using Archimedes’ principle, relate the buoyant force on the ball to its diameter:

3w6

1ballwff dgVgmwB πρρ ====

Substitute to obtain: 03w6

1 =−− Tmgdπρ

Solve for d: ( )

3

w

6πρ

mgTd +=

Substitute numerical values and evaluate d:

( )( )[ ]( ) cm05.5

kg/m10m/s9.81kg0.004N108.26

333

22

=+×

=−

πd

Chapter 13

1022

73 • Picture the Problem Let ρ0 represent the density of seawater at the surface. We can use the definition of density and the fact that mass is constant to relate the fractional change in the density of water to its fractional change in volume. We can also use the definition of bulk modulus to relate the fractional change in density to the increase in pressure with depth and solve the resulting equation for the density at the depth at which the pressure is 800 atm. Using the definition of density, relate the mass of a given volume of seawater to its volume:

Vm ρ=

Noting that the mass does not vary with depth, evaluate its differential:

0=+ ρρ VddV

Solve for dρ/ρ: VdVd

−=ρρ

orVV∆

−≈∆ρρ

Using the definition of the bulk modulus, relate ∆P to ∆ρ/ρ0: 0ρρ∆

∆=

∆∆

−=P

VVPB

Solve ∆ρ:

BP∆

=−=∆ 00

ρρρρ

Solve for ρ:

⎟⎠⎞

⎜⎝⎛ ∆+=

∆+=

BP

BP 10

00 ρρρρ

Substitute numerical values and evaluate ρ:

( ) 329

5

3 kg/m1061N/m102.3

atm1Pa101.01atm800

1kg/m1025 =⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

×

××

+=ρ

Fluids

1023

74 • Picture the Problem When it is submerged, the block is in equilibrium under the influence of the buoyant force due to the water, the force exerted by the spring balance, and its weight. We can use the condition for translational equilibrium to relate the buoyant force to the weight of the block and the definition of density to express the weight of the block in terms of its density. Apply ∑ = 0yF to the block: 08.0 =−+ mgmgB ⇒ mgB 2.0=

Substitute for B and m to obtain: gVgV blockblockblockw 2.0 ρρ =

Solve for and evaluate ρblock: ( )

33

33wblock

kg/m105.00

kg/m1052.0

×=

==ρρ

*75 • Picture the Problem When the copper block is floating on a pool of mercury, it is in equilibrium under the influence of its weight and the buoyant force acting on it. We can apply the condition for translational equilibrium to relate these forces. We can find the fraction of the block that is submerged by applying Archimedes’ principle and the definition of density to express the forces in terms of the volume of the block and the volume of the displaced mercury. Let V represent the volume of the copper block, V′ the volume of the displaced mercury. Then the fraction submerged when the material is floated on water is V′/ V. Choose the upward direction to be the positive y direction. Apply ∑ = 0yF to the block:

0=− wB , where B is the buoyant force and w is the weight of the block.

Apply Archimedes’ principle and the definition of density to obtain:

0CuHg =− VgV'g ρρ

Solve for V′/ V:

Hg

Cu

ρρ

=VV'

Chapter 13

1024

Substitute numerical values and evaluate V′/ V:

%7.65657.0kg/m106.13kg/m1093.8

33

33

==××

=VV'

76 • Picture the Problem When the block is floating on a pool of ethanol, it is in equilibrium under the influence of its weight and the buoyant force acting on it. We can apply the condition for translational equilibrium to relate these forces. We can find the fraction of the block that is submerged by applying Archimedes’ principle and the definition of density to express the forces in terms of the volume of the block and the volume of the displaced ethanol. Let V represent the volume of the copper block, V′ the volume of the displaced ethanol. Then the fraction of the volume of the block that will be submerged when the material is floated on water is V′/ V. Choose the upward direction to be the positive y direction. Apply ∑ = 0yF to the block

floating on ethanol:

0eth =− wB , where Beth is the buoyant

force due to the ethanol and w is the weight of the block.


( )gVw 9.0ethρ=

Apply ∑ = 0yF to the block

floating on water:

0w =− wB , where Bw is the buoyant force

due to the water and w is the weight of the block.


V'gw wρ= , where V′ is the volume of the

displaced water.

Equate the two expressions for w and solve for V′/ V: w

eth9.0ρρ

=VV'

Substitute numerical values and evaluate V′/ V:

( )

%5.72725.0

kg/m10kg/m10806.09.033

33

==

×=

VV'

77 • Determine the Concept If you are floating, the density (or specific gravity) of the liquid in which you are floating is immaterial as you are in translational equilibrium under the influence of your weight and the buoyant force on your body. Thus, the buoyant force on your body is your weight in both (a) and (b).

Fluids

1025

78 • Picture the Problem Let m and V represent the mass and volume of your body. Because you are in translational equilibrium when you are floating, we can apply the condition for translational equilibrium and Archimedes’ principle to your body to express the dependence of the volume of water it displaces when it is fully submerged on your weight. Let the upward direction be the positive y direction. Apply ∑ = 0yF to your floating body:

0=− mgB

Use Archimedes’ principle to relate the density of water to your volume:

( )gVgmwB 96.0wff ρ===

Substitute to obtain: ( ) 096.0w =− mggVρ

Solve for V:

w96.0 ρmV =

79 •• Picture the Problem Let m and V represent the mass and volume of the block of wood. Because the block is in translational equilibrium when it is floating, we can apply the condition for translational equilibrium and Archimedes’ principle to express the dependence of the volume of water it displaces when it is fully submerged on its weight. We’ll repeat this process for the situation in which the lead block is resting on the wood block with the latter fully submerged. Let the upward direction be the positive y direction. Apply ∑ = 0yF to floating block: 0=− mgB

Use Archimedes’ principle to relate the density of water to the volume of the block of wood:

( )gVgmwB 68.0wff ρ===

Using the definition of density, express the weight of the block in terms of its density:

Vgmg woodρ=

Substitute to obtain: ( ) 068.0 woodw =− VggV ρρ

Solve for and evaluate the density of the wood block:

( )3

33wwood

kg/m680

kg/m1068.068.0

=

== ρρ

Chapter 13

1026

Use the definition of density to find the volume of the wood:

33

3wood

wood

m102.206

kg/m680kg1.5

−×=

==ρmV

Apply ∑ = 0yF to the floating

block when the lead block is placed on it:

0' =− gmB' , where B′ is the new

buoyant force on the block and m′ is the combined mass of the wood block and the lead block.

Use Archimedes’ principle and the definition of density to obtain:

( ) 0blockPbw =+− gmmVgρ

Solve for the mass of the lead block:

blockwPb mVm −= ρ


( )( )

kg0.706

kg1.5m102.206kg/m10 3333

Pb

=

−×= −m

*80 •• Picture the Problem The true mass of the Styrofoam cube is greater than that indicated by the balance due to the buoyant force acting on it. The balance is in rotational equilibrium under the influence of the buoyant and gravitational forces acting on the Styrofoam cube and the brass masses. Neglect the buoyancy of the brass masses. Let m and V represent the mass and volume of the cube and L the lever arm of the balance. Apply 0=∑ τr to the balance:

( ) 0brass =−− gLmLBmg

Use Archimedes’ principle to express the buoyant force on the Styrofoam cube as a function of volume and density of the air it displaces:

VgB airρ=

Substitute and simplify to obtain: 0brassair =−− mVm ρ

Solve for m: brassair mVm += ρ


( )( )g40.2kg104.02

kg1020m0.25kg/m1.2932

333

=×=

×+=−

−m

Fluids

1027

81 •• Picture the Problem Let din and dout represent the inner and outer diameters of the copper shell and V′ the volume of the sphere that is submerged. Because the spherical shell is floating, it is in translational equilibrium and we can apply a condition for translational equilibrium to relate the buoyant force B due to the displaced water and its weight w. Apply ∑ = 0yF to the spherical shell:

0=− wB

Using Archimedes’ principle and the definition of w, substitute to obtain:

0w =−mgV'gρ

or 0w =−mV'ρ (1)

Express V′ as a function dout: 3

out3out 1262

1 ddV' ππ==

Express m in terms of din and dout: ( )

⎟⎠⎞

⎜⎝⎛ −=

−=

3in

3outCu

inoutCu

66dd

VVmππρ

ρ


66123in

3outCu

3outw =⎟

⎠⎞

⎜⎝⎛ −− ddd ππρπρ

Simplify: ( ) 0

21 3

in3outCu

3outw =−− ddd ρρ

Solve for din:

3

Cu

woutin 2

1ρρ

−= dd

Substitute numerical values and evaluate din:

( ) ( ) cm8.1193.82

11cm12 3in =−=d

82 •• Determine the Concept The additional weight on the beaker side equals the weight of the displaced water, i.e., 64 g. This is the mass that must be placed on the other cup to maintain balance. *83 •• Picture the Problem We can use the definition of Reynolds number and assume a value for NR of 1000 (well within the laminar flow range) to obtain a trial value for the radius

Chapter 13

1028

of the pipe. We’ll then use Poiseuille’s law to determine the pressure difference between the ends of the pipe that would be required to maintain a volume flow rate of 500 L/s. Use the definition of Reynolds number to relate NR to the radius of the pipe:

ηρvrN 2

R =

Use the definition of IV to relate the volume flow rate of the pipe to its radius:

vrAvI 2V π== ⇒ 2

V

rIvπ

=


INηπρ V

R2

=

Solve for r:

R

V2NIr

ηπρ

=

Substitute numerical values and evaluate r: ( )( )

( )( ) cm9.271000sPa8.0

/sm0.500kg/m7002 33

=⋅

=π

r

Using Poiseuille’s law, relate the pressure difference between the ends of the pipe to its radius:

V48 I

rLP

πη

=∆


( )( )( )

( )

atm0.83Pa101.01325

atm1Pa1041.8

Pa1041.8

/sm500.0m279.0

km50sPa0.88

56

6

34

=×

××=

×=

⋅=∆

πP

This pressure is too large to maintain in the pipe.

Evaluate ∆P for a pipe of 50 cm radius: ( )( )( )

( )

atm04.8Pa101.01325

atm1Pa1015.8

Pa1015.8

/sm500.0m50.0

km50sPa0.88

55

5

34

=×

××=

×=

⋅=∆

πP

Fluids

1029

pipeline. for thediameter reasonable a is m 1

84 •• Picture the Problem We’ll measure the height of the liquid–air interfaces relative to the centerline of the pipe. We can use the definition of the volume flow rate in a pipe to find the speed of the water at point A and the relationship between the gauge pressures at points A and C to determine the level of the liquid-air interface at A. We can use the continuity equation to express the speed of the water at B in terms of its speed at A and Bernoulli’s equation for constant elevation to find the gauge pressure at B. Finally, we can use the relationship between the gauge pressures at points A and B to find the level of the liquid-air interface at B. Relate the gauge pressure in the pipe at A to the height of the liquid-air interface at A:

AAgauge, ghP ρ=

where hA is measured from the center of the pipe.

Solve for hA: g

Ph

ρAgauge,

A =

Substitute numerical values and evaluate hA:

( )( )( )( )

m6.12

m/s9.81kg/m10Pa/atm101.01atm1.22

233

5

A

=

×=h

Determine the velocity of the water at A: ( )

m/s55.2m02.0

4

/sm108.02

33

A

VA =

×==

−

πAIv

Apply Bernoulli’s equation for constant elevation to relate PB and PA:

2B2

1B

2A2

1A vPvP ρρ +=+ (1)

Use the continuity equation to relate vB and vA:

BBAA vAvA =

Solve for vB: ( )( ) AA2

2

A2B

2A

AB

AB 4

cm1cm2 vvv

ddv

AAv ====


AB2A2

1A 8 vPvP ρρ +=+

Chapter 13

1030

Solve for PB: 2A2

15AB vPP ρ−=

Substitute numerical values and evaluate PB:

( )( )( )( )

atm733.1Pa101.01

atm1Pa1075.1

Pa1075.1

m/s2.55kg/m10

Pa/atm1001.1atm22.2

55

5

2332

15

5B

=×

××=

×=

−

×=P

Relate the gauge pressure in the pipe at B to the height of the liquid-air interface at B:

BBgauge, ghP ρ=

Solve for hB: g

Ph

ρBgauge,

B =

Substitute numerical values and evaluate hB:

( )[ ]( )( )m55.7

m/s9.81kg/m10atmPa101.01atm11.733

233

5

B

=

⎟⎠⎞

⎜⎝⎛ ×−

=h

85 •• Picture the Problem We’ll measure the height of the liquid–air interfaces relative to the centerline of the pipe. We can use the definition of the volume flow rate in a pipe to find the speed of the water at point A and the relationship between the gauge pressures at points A and C to determine the level of the liquid-air interface at A. We can use the continuity equation to express the speed of the water at B in terms of its speed at A and Bernoulli’s equation for constant elevation to find the gauge pressure at B. Finally, we can use the relationship between the gauge pressures at points A and B to find the level of the liquid-air interface at B. Relate the gauge pressure in the pipe at A to the height of the liquid-air interface at A:

AAgauge, ghP ρ=

where hA is measured from the center of the pipe.

Solve for hA: g

Ph

ρAgauge,

A =

Fluids

1031

Substitute numerical values and evaluate hA:

( )( )( )( )

m6.12

m/s9.81kg/m10Pa/atm101.01atm1.22

233

5

A

=

×=h

Determine the velocity of the water at A: ( )

m/s91.1m02.0

4

/sm106.02

33

A

VA =

×==

−

πAIv

Use the continuity equation to relate vB and vA:

BBAA vAvA =

Solve for vB: ( )( )

A

A2

2

A2B

2A

AB

AB

4cm1cm2

v

vvddv

AAv

=

===

Apply Bernoulli’s equation for constant elevation to relate PB and PA:

2B2

1B

2A2

1A vPvP ρρ +=+ (1)

Substitute in equation (1) to obtain: 2AB

2A2

1A 8 vPvP ρρ +=+

Solve for PB: 2

A215

AB vPP ρ−=

Substitute numerical values and evaluate PB:

( )( )( )( )

atm95.1Pa101.01

atm1Pa10969.1

Pa10969.1

m/s91.1kg/m10

Pa/atm1001.1atm22.2

55

5

2332

15

5B

=×

××=

×=

−

×=P

Relate the gauge pressure in the pipe at B to the height of the liquid-air interface at B:

BBgauge, ghP ρ=

Solve for hB: g

Ph

ρBgauge,

B =

Chapter 13

1032

Substitute numerical values and evaluate hB:

( )[ ]( )( )

m78.9

m/s9.81kg/m10atmPa101.01atm11.95

233

5

B

=

⎟⎠⎞

⎜⎝⎛ ×−

=h

*86 •• Picture the Problem Because it is not given, we’ll neglect the difference in height between the centers of the pipes at A and B. We can use the definition of the volume flow rate to find the speed of the water at A and Bernoulli’s equation for constant elevation to find its speed at B. Once we know the speed of the water at B, we can use the equation of continuity to find the diameter of the constriction at B. Use the definition of the volume flow rate to find vA: ( )

m/s59.1m02.0

4

/sm105.02

33

A

VA =

×==

−

πAIv

Use Bernoulli’s equation for constant elevation to relate the pressures and velocities at A and B:

2A2

1A

2B2

1B vPvP ρρ +=+

Solve for 2Bv :

( ) 2A

BA2B

2 vPPv +−

=ρ

Substitute numerical values and evaluate 2

Bv :

( ) ( )[ ] ( ) 222

33

52B /sm222m/s1.59

kg/m10Pa/atm101.01atm0.11.1872

=+×−

=v

Using the continuity equation, relate the volume flow rate to the radius at B:

B2

BBBV vrvAI π==

Solve for and evaluate rB:

( ) mm27.3m/s14.9

/sm100.5 33

B

VB =

×==

−

πvIrπ

and mm54.62 BB == rd

Fluids

1033

87 •• Picture the Problem Let V′ represent the volume of the buoy that is submerged and h′ the height of the submerged portion of the cylinder. We can find the fraction of the cylinder’s volume that is submerged by applying the condition for translational equilibrium to the buoy and using Archimedes’ principle. When the buoy is submerged it is in equilibrium under the influence of the tension T in the cable, the buoyant force due to the displaced water, and its weight. When the cable breaks, the net force acting on the buoy will accelerate it and we can use Newton’s 2nd law to find its acceleration. (a) Apply ∑ = 0yF to the cylinder: 0=− wB

Using Archimedes’ principle and the definition of weight, substitute for B and w:

0sw =−mgV'gρ

or 0sw =−mgh'Agρ

where A is the cross-sectional area of the buoy.

Solve for and evaluate h′: A

mh'swρ

=

Substitute numerical values and evaluate h′: ( ) ( )

m920.0

m0.94πkg/m101.025

kg600233

=

×=h'

Use h′ to find the height h of the buoy:

m1.68m0.920m6.2 =−=− h'h

Express the fraction of the volume of the cylinder that is above water:

hh'

hd

h'd

VV'

VV'V

−=

−=−=−

1

4

4112

2

π

π

Substitute numerical values to obtain: %6.64

m2.6m0.9201 =−=

−V

V'V

(b) Apply ∑ = 0yF to the

submerged buoy:

0=−− wTB

Chapter 13

1034

Solve for T and substitute for B and w to obtain: ( )gmV

mgVgwBT−=

−=−=

sw

sw

ρρ

Substitute numerical values and evaluate T: ( )[ ( ) ( )

]( )kN7.10

m/s9.81kg600

m2.6m0.94

kg/m101.025

2

233

=

−

×=πT

(c) Apply ∑ = 0yF to the buoy: mawB =−

Substitute for B − w and solve for a to obtain: m

Tm

wBa =−

=

Substitute numerical values and evaluate a: 2m/s17.9

kg600kN10.75

==a

88 •• Picture the Problem Because the floating object is in equilibrium under the influence of the buoyant force acting on it and its weight; we can apply the condition for translational equilibrium to relate B and w. Let ∆h represent the change in elevation of the liquid level and Vf the volume of the displaced fluid. Apply ∑ = 0yF to the floating

object:

0=− wB

Using Archimedes’ principle and the definition of weight, substitute for B and w:

0f0 =−mggVρ

The volume of fluid displaced is the sum of the volume displaced in the two vessels:

hAhAhAVVV AA

∆=∆+∆=∆+∆=

433f

Substitute for Vf to obtain: 04 0 =−∆ mghgAρ

Solve for ∆h:

04 ρAmh =∆

Fluids

1035

89 •• Picture the Problem We can calculate the smallest pressure change ∆P that can be detected from the reading ∆h from .hgP ∆=∆ ρ

Express and evaluate the pressure difference between the two columns of the manometer:

( )( )( )

Pa4415.0m100.05

m/s9.81kg/m9003

23

=××

=

∆=∆

−

hgP ρ

Express this pressure in mmHg and µmHg:

mHg31.3

mmHg1031.3

atm1mmHg760

Pa101.01325atm1Pa4415.0

3

5

µ=

×=

×

××=∆

−

P

90 •• Picture the Problem We can use the equality of the pressure at the bottom of the U-tube due to the water on one side and that due to the oil and water on the other to relate the various heights. Let h represent the height of the oil above the water. Then ho = h1w + h.

Using the constancy of the amount of water, express the relationship between h1w and h2w:

h1w + h2w = 56 cm

Find the height of the oil-water interface:

cm22.0cm34cm651w =−=h

Express the equality of the pressure at the bottom of the two arms of the U tube:

( ) ( ) oilwww 78.0cm22cm34 ghgg ρρρ +=

Chapter 13

1036

Solve for and evaluate hoil: ( ) ( )

( ) ( ) cm4.1578.0

cm22cm3478.0

cm22cm34

w

wwoil

=−

=

−=

gggh

ρρρ

Find the height of the air-oil interface ho:

cm37.4cm15.422cmo =+=h

91 •• Picture the Problem Let σL represent the specific gravity of the liquid. The specific gravity of the oil is σo = 0.8. We can use the equality of the pressure at the bottom of the U-tube due to the water on one side and that due to the oil and water on the other to relate the various heights.

Express the equality of the pressure at the bottom of the two arms of the U tube:

( ) ( )cm128.0cm7 wLL ghggh σσσ +−=

Solve for and evaluate σL: ( ) ( )

37.1

cm7cm128.0

cm7cm128.0 w

L

=

==σ

σ

92 •• Picture the Problem The block of wood is in translational equilibrium under the influence of the buoyant force due to the displaced water acting on it and on the lead block, its weight, and the weight of the lead block. We can use a condition for translational equilibrium and Archimedes’ principle to obtain a relationship between the mass of the lead block and the densities of water, wood, and lead and the mass of the wood block. Apply ∑ = 0yF to the block of wood:

0PbwoodPbwood =−−+ gmgmBB

Use Archimedes’ principle to express the buoyant force on the block of wood:

gVB woodwwood ρ=

Use Archimedes’ principle to gVB PbwPb ρ=

Fluids

1037

express the buoyant force on the lead block:

Substitute and simplify to obtain: 0PbwoodPbwwoodw =−−+ mmVV ρρ

Express the volume of the wood block in terms of its density and mass:

wood

woodwood ρ

mV =

Express the volume of the lead block in terms of its density and mass:

Pb

PbPb ρ

mV =

Substitute for Vwood and VPb: 0Pbwood

Pb

Pbw

wood

woodw =−−+ mmmm

ρρ

ρρ

Solve for mPb:

Pb

w

woodwood

w

Pb

1

1

ρρ

ρρ

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=m

m


( )kg235.0

3.1111

kg5.017.0

1

Pb =−

⎟⎠⎞

⎜⎝⎛ −

=m

*93 •• Picture the Problem Because the balloon is in equilibrium under the influence of the buoyant force exerted by the air, the weight of its basket and load w, the weight of the skin of the balloon, and the weight of the helium. Choose upward to be the positive y direction and apply the condition for translational equilibrium to relate these forces. Archimedes’ principle relates the buoyant force on the balloon to the density of the air it displaces and the volume of the balloon. (a) Apply ∑ = 0yF to the balloon: 0Heskin =−−− wgmgmB

Letting V represent the volume of the balloon, use Archimedes’ principle to express the buoyant force:

0Heskinair =−−− wgmgmVgρ

Substitute for mHe: 0Heskinair =−−− wVggmVg ρρ

Chapter 13

1038

Solve for V: ( )g

wgmVHeair

skin

ρρ −+

=


( )( )( )( )( )

3

23

2

m0.70

m/s9.81kg/m0.17861.293N750m/s9.81kg1.5

=

−+

=V

(b) Apply ∑ = maFy to the balloon: amgmB tottot =−

Solve for a: g

mBa −=tot

Assuming that the mass of the skin has not changed and letting V′ represent the doubled volume of the balloon, express mtot:

skinHeload

skinHeloadtot

' mVg

wmmmm

++=

++=

ρ

Substitute numerical values and evaluate mtot:

( )( ) kg118kg1.5m140kg/m0.1786m/s9.81N900 33

2tot =++=m

Express the buoyant force acting on the balloon:

gVwB 'airfluid displaced ρ==

Substitute numerical values and evaluate B:

( )( )( )kN78.1

m/s9.81m140kg/m1.293 233

==B

Substitute and evaluate a: 22 m/s27.5m/s9.81kg118kN1.78

=−=a

94 •• Picture the Problem When the hollow sphere is completely submerged but floating, it is in translational equilibrium under the influence of a buoyant force and its weight. The buoyant force is given by Archimedes’ principle and the weight of the sphere is the sum of the weights of the hollow sphere and the material filling its center. Apply ∑ = 0yF to the hollow sphere: 0=−wB

Fluids

1039

Express the buoyant force acting on the hollow sphere:

( )[ ]gR

gRgVB3

0364

334

0sphere0 222

πρ

πρρ

=

==

Express the weight of the sphere when it’s hollow is filled with a material of density ρ′:

( ) [ ] [ ]gR'gR

gR'gRR

gVgVw

3343

0328

33433

34

0

hollowspherehollow0

2

'

πρπρ

πρπρ

ρρ

+=

+−=

+=


343

03283

0364 =−− gR'gRgR πρπρπρ

Solve for ρ′:

09' ρρ =

*95 •• Picture the Problem We can differentiate the function P(h) to show that it satisfies the differential equation dP/P = −C dh and in part (b) we can use the approximation e−x ≈ 1 – x and ∆h << h0 to establish the given result. (a) Differentiate P(h) = P0 e−Ch:

P

ePdhdP Ch

C

C 0

−=

−= −

Separate variables to obtain:

dhP

dP C−=

(b) Express P(h + ∆h): ( ) ( )

( ) h

hh

hh

ehP

eeP

ePhhP

∆−

∆−−

∆+−

=

=

=∆+

C

CC0

C0

For ∆h << h0: 1

0

<<∆hh

Let h0 = 1/C. Then: 1C <<∆h

and

0

C 1C1hhhe h ∆

−=∆−≈∆−


( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛ ∆−=∆+

0

1hhhPhhP

Chapter 13

1040

(c) Take the logarithm of both sides of the function P(h): hP

ePePP hh

Clnlnlnlnln

0

C0

C0

−=+== −−

Solve for C:

⎟⎠⎞

⎜⎝⎛=

PP

h0ln1C

Substitute numerical values and evaluate C:

1

021

0

km126.0

2lnkm5.5

1lnkm5.5

1C

−=

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

PP

96 •• Picture the Problem Let V represent the volume of the submarine and V ′ the volume of seawater it displaces when it is on the surface. The submarine is in equilibrium in both parts of the problem. Hence we can apply the condition for translational equilibrium (neutral buoyancy) to the submarine to relate its weight to the buoyant force acting on it. We’ll also use Archimedes’ principle to connect the buoyant forces to the volume of seawater the submarine displaces. Let upward be the positive y direction. (a) Express f, the fraction of the submarine’s volume above the surface when the tanks are filled with air:

VV'

VV'Vf −=

−= 1 (1)

Apply ∑ = 0yF to the submarine

when its tanks are full of air:

0=− wB

Use Archimedes’ principle to express the buoyant force on the submarine in terms of the volume of the displaced water:

V'gB swρ=

Substitute and solve for V′:

swρmV' =


Vmfsw

1ρ

−=

Fluids

1041

Substitute numerical values and evaluate f: ( )( )

%44.21044.2

m102.4kg/m101.025kg102.41

2

3333

6

=×=

×××

−=

−

f

(b) Express the volume of seawater in terms of its mass and density: sw

swsw ρ

mV = (2)

Apply ∑ = 0yF , the condition for

neutral buoyancy, to the submarine:

0swsub =−− wwB

Use Archimedes’ principle to express the buoyant force on the submarine in terms of the volume of the displaced water:

VgB swρ=

Substitute to obtain: 0swsubsw =−− gmgmVgρ

Solve for msw: subswsw mVm −= ρ

Substitute for Vsw in equation (2) to obtain: sw

sub

sw

subswsw ρρ

ρ mVmVV −=−

=

Substitute numerical values and evaluate Vsw:

3

33

633

sw

m5.58

kg/m101.025kg102.4m102.4

=

××

−×=V

97 •• Picture the Problem While the loaded crate is under the surface, it is in equilibrium under the influence of the tension in the cable, the buoyant force acting on the gold, and the gravitational force acting on the gold. The empty crate has neutral buoyancy. When the crate is out of the water, the buoyant force of the air is negligible and the tension in the cable is the sum of the weights of the crate, the gold bullion, and the seawater. (a) Apply ∑ = 0yF to the crate

while it is below the surface:

0AuAu =−+ wBT

Solve for the tension in the cable:

AuAu BwT −=

Using Archimedes’ principle, relate the buoyant force acting on the gold

gVB AuswAu ρ=

Chapter 13

1042

to its density and volume: Substitute for BAu and simplify to obtain:

( ) gVT AuswAu ρρ −=


( )[ ( )]( )( )( )( )( )kN9.33

m/s9.81m0.5m0.75m1.4.360kg/m101.025kg/m1019.3 23333

=

×−×=T

(b) 1. Apply ∑ = 0yF to the crate

while it is being lifted to the deck of the ship with none of the seawater leaking out:

0swcrateAu =−−− wwwT

Substitute for the weights of the gold, crate, and seawater and solve for the tension in the cable and express: ( )gVmV

gVgmgVwwwT

swswcrateAuAu

swswcrateAuAu

swcrateAu

ρρρρ

++=++=

++=


( )( )( )( )( )[ ( )( )( )( )( )]( )

kN8.39

m/s9.81m0.5m0.75m1.40.64kg/m101.025kg32m0.5m0.75m1.40.36kg/m1019.3

2

3333

=

×

×++×=T

2. With the seawater term missing, the expression for the tension is:

( )gmVgmgV

wwT

crateAuAu

crateAuAu

crateAu

+=+=

+=

ρρ


( )( )( )( )( )[ ]( ) kN1.36m/s9.81kg32m0.5m0.75m1.40.36kg/m1019.3 233 =+×=T

98 ••• Picture the Problem In the three situations described in the problem the hydrometer will be in equilibrium under the influence of its weight and the buoyant force exerted by the liquids. We can use Archimedes’ principle to relate the buoyant force acting on the hydrometer to the density of the liquid in which it is floating and to its weight.

Fluids

1043

(a) Find the volume of the bulb: ( ) 33613

61

bulb cm238.7cm4.2 === ππdV

Find the volume of the tube: ( ) ( )

2

2412

41

tube

cm836.8

cm20cm75.0

=

== ππ LdV

Apply ∑ = 0yF to the hydrometer

just floating in the liquid:

0Pbhyd =−− gmwB

Substitute for B and wglass: 0Pbglasshydliq =−− gmgmgVρ

Solve for mPb: hydhydliqPb mVm −= ρ


( )( )

g26.5

g28.7cm8.836cm7.238

g/cm178.033

3Pb

=

−+×

=m

(b) Letting V represent the volume of the hydrometer that is submerged, apply ∑ = 0yF to the hydrometer

just floating in the liquid:

0w =− mgVgρ

Solve for V:

w

Pbhyd

w ρρmmmV

+==


33 cm54.12

g/cm15.26g7.28

=+

=V

Relate the volume of the hydrometer that is submerged to the volume of the bulb and the volume of the tube that is submerged:

bulb2tube4

1 Vh'dV += π

Solve for h′: 2tube4

1bulb

dVVh'

π−

=

Substitute numerical values and evaluate h′: ( )

cm0.12cm75.0

cm238.7cm54.122

41

33

=−

=π

h'

Chapter 13

1044

Find the length of the tube that shows above the surface of the water: cm8.00

cm12.0cm20'cm20

=

−=−= hh

(c) Apply ∑ = 0yF to the

hydrometer floating in the liquid of unknown specific gravity:

0hydLL =− gmgVρ

Solve for the density of the liquid:

L

hydL V

m=ρ

Express the volume of the displaced liquid:

'2tube4

1bulbL hdVV π+=

Substitute numerical values and evaluate VL:

( )( )

3

2413

L

cm68.10cm12.2cm20

cm75.0cm238.7

=

−×

+= πV

Substitute for VL and mhyd and evaluate ρL:

33L g/cm174.1

cm10.68g12.54

==ρ

Express and evaluate the specific gravity of the liquid:

17.1gravityspecificL

wliquid ==

ρρ

99 ••• Picture the Problem We can apply Bernoulli’s equation to the top of the keg and to the spigot opening to determine the rate at which the root beer exits the tank. Because the area of the spigot is much smaller than that of the keg, we can neglect the velocity of the root beer at the top of the keg. We’ll use the continuity equation to obtain an expression for the rate of change of the height of the root beer in the keg as a function of the its height and integrate this function to find h as a function of time. (a) Apply Bernoulli’s equation to the beer at the top of the keg and at the spigot:

22beer2

1

2beer221beer2

1beer1

v

ghPvghP

ρ

ρρρ

+

+=++

or, because v1 ≈ 0, h2 = 0, P1 = P2 = Pat, and h1 = h,

222

1 vgh =

Solve for v2: ghv 22 =

Fluids

1045

(b) Use the continuity equation to relate v1 and v2:

2211 vAvA =

Substitute −dh/dt for v1: 221 vA

dtdhA =−

Substitute for v2 and solve for dh/dt to obtain:

ghAA

dtdh 2

1

2−=

(c) Separate the variables in the differential equation:

dth

dhgAA

=−2

21

Express the integral from h′ = H to h and t′ = 0 to t: ∫∫ =−

th

H

dt'h

dh'gAA

0

21

'2

Evaluate the integral to obtain: ( ) thH

gAA

=−−2

21

Solve for h: 2

1

2 22 ⎟⎟

⎠

⎞⎜⎜⎝

⎛−= tg

AAHh

(d) Solve h(t) for the time-to-drain t′: g

HAAt' 2

2

1=

Substitute numerical values and evaluate t′

( )

min 46h 1

s1039.6m/s81.9m22

103

21

41

=

×== − AAt'

Chapter 13

1046

1047

Chapter 14 Oscillations Conceptual Problems 1 • Determine the Concept The acceleration of an oscillator of amplitude A and frequency f is zero when it is passing through its equilibrium position and is a maximum when it is at its turning points.

When v = vmax: 0=a

When x = xmax: AfAa 222 4πω ==

2 • Determine the Concept The condition for simple harmonic motion is that there be a linear restoring force; i.e., that F = −kx. Thus, the acceleration and displacement (when they are not zero) are always oppositely directed. v and a can be in the same direction, as can v and x.

3 • (a) False. In simple harmonic motion, the period is independent of the amplitude. (b) True. In simple harmonic motion, the frequency is the reciprocal of the period which, in turn, is independent of the amplitude. (c) True. The condition that the acceleration of a particle is proportional to the displacement and oppositely directed is equivalent to requiring that there be a linear restoring force; i.e., F = −kx ⇔ ma = −kx or a = − (k/m)x. *4 • Determine the Concept The energy of a simple harmonic oscillator varies as the square of the amplitude of its motion. Hence, tripling the amplitude increases the energy by a factor of 9. 5 •• Picture the Problem The total energy of an object undergoing simple harmonic motion is given by ,2

21

tot kAE = where k is the stiffness constant and A is the amplitude of the

motion. The potential energy of the oscillator when it is a distance x from its equilibrium position is ( ) .2

21 kxxU =

Chapter 14

1048

Express the ratio of the potential energy of the object when it is 2 cm from the equilibrium position to its total energy:

( )2

2

221

221

tot Ax

kAkx

ExU

==

Evaluate this ratio for x = 2 cm and A = 4 cm:

( ) ( )( ) 4

1cm4cm2cm2

2

2

tot

==E

U

( ) ( )

( ) 41

cm4cm2cm2

2

2

tot

==E

U

and correct. is )(a

6 • (a) True. The factors determining the period of the object, i.e., its mass and the spring constant, are independent of the oscillator’s orientation. (b) True. The factors determining the maximum speed of the object, i.e., its amplitude and angular frequency, are independent of the oscillator’s orientation. 7 • False. In order for a simple pendulum to execute simple harmonic motion, the restoring force must be linear. This condition is satisfied, at least approximately, for small initial angular displacements. 8 • True. In order for a simple pendulum to execute periodic motion, the restoring force must be linear. This condition is satisfied for any initial angular displacement. *9 •• Determine the Concept Assume that the first cart is given an initial velocity v by the blow. After the initial blow, there are no external forces acting on the carts, so their center of mass moves at a constant velocity v/2. The two carts will oscillate about their center of mass in simple harmonic motion where the amplitude of their velocity is v/2. Therefore, when one cart has velocity v/2 with respect to the center of mass, the other will have velocity −v/2. The velocity with respect to the laboratory frame of reference will be +v and 0, respectively. Half a period later, the situation is reversed; one cart will move as the other stops, and vice-versa. *10 •• Determine the Concept The period of a simple pendulum depends on the reciprocal of the length of the pendulum. Increasing the length of the pendulum will decrease its period

Oscillations

1049

and the clock would run slow. 11 • True. The mechanical energy of a damped, undriven oscillator varies with time according to τteEE −= 0 where E0 is the oscillator’s energy at t = 0 and τ is the time constant. 12 • (a) True. The amplitude of the motion of a driven oscillator depends on the driving (ω)

and natural (ω0) frequencies according to ( ) .222220

20 ωωω bmFA +−= When

ω = ω0, the amplitude of the motion is a maximum and is given by .220 ωbFA =

(b) True. The width of the resonance curve (∆ω) depends on the Q value according to

Q10 =∆ ωω . Thus when Q is large, ∆ω is small and the resonance is sharp.

13 • Determine the Concept Examples of driven oscillators include the pendulum of a clock, a bowed violin string, and the membrane of any loudspeaker.

14 • Determine the Concept The shattering of a crystal wineglass is a consequence of the glass being driven at or near its resonant frequency. correct. is )(a

*15 • Determine the Concept We can use the expression for the frequency of a spring-and-mass oscillator to determine the effect of the mass of the spring.

If m represents the mass of the object attached to the spring in a spring-and-mass oscillator, the frequency is given by:

mkf

π21

=

If the mass of the spring is taken into account, the effective mass is greater than the mass of the object alone.

eff21

mkf'

π=

Divide the second of these equations by the first and simplify to obtain:

eff

eff

21

21

mm

mk

mk

ff'

==

π

π

Chapter 14

1050

Solve for f ′:

effmmff' =

reduced. be willfrequency that thepredicts spring theof mass effective eaccount th into taking ofroot square thersely withinv varies Because m,ef'

16 •• Determine the Concept The period of the lamp varies inversely with the square root of the effective value of the local gravitational field.

1. greater than T0 when B. the train rounds a curve of radius R with

speed v.

2. less than T0 when D. the train goes over the crest of a hill of radius of curvature R with constant speed.

3. equal to T0 when A. the train moves horizontally with constant velocity.

C. the train climbs a hill of inclination θ at constant speed.

17 ••

Picture the Problem We can use Mkf

π21

= to express the frequencies of the two

mass-spring systems in terms of their masses. Dividing one of the equations by the other will allow us to express MA in terms of MB.

Express the frequency of mass-spring system A as a function of its mass:

AA 2

1Mkf

π=

Express the frequency of mass-spring system B as a function of its mass:

BB 2

1Mkf

π=


B

A

A

B

MM

ff

=

Solve for MA: B4

1B

2

B

BB

2

A

BA 2

MMf

fMffM =⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

Oscillations

1051

and correct. is )(d

18 •• Picture the Problem We can relate the energies of the two mass-spring systems through either 2

21 kAE = or 22

21 AME ω= and investigate the relationship between their

amplitudes by equating the expressions, substituting for MA, and expressing AA in terms of AB.

Express the energy of mass-spring system A:

2A

2AA2

12AA2

1A AMAkE ω==

Express the energy of mass-spring system B:

2B

2BB2

12BB2

1B AMAkE ω==


2B

2BB2

1

2A

2AA2

1

B

A 1AMAM

EE

ωω

==

Substitute for MA and simplify:

2B

2B

2A

2A

2B

2BB

2A

2AB 221

AA

AMAM

ωω

ωω

==

Solve for AA:

BA

BA 2

AAω

ω=

Without knowing how ωA and ωB, or kA and

kB, are related, we cannot simplify this expression further. correct. is )(d

19 •• Picture the Problem We can express the energy of each system using 2

21 kAE = and,

because the energies are equal, equate them and solve for AA. Express the energy of mass-spring system A in terms of the amplitude of its motion:

2AA2

1A AkE =

Express the energy of mass-spring system B in terms of the amplitude of its motion:

2BB2

1B AkE =

Chapter 14

1052

Because the energies of the two systems are equal we can equate them to obtain:

2BB2

12AA2

1 AkAk =

Solve for AA: B

A

BA A

kkA =

Substitute for kA and simplify to obtain: 22

BB

B

BA

AAk

kA ==

and correct. is )(b

20 •• Picture the Problem The period of a simple pendulum is independent of the mass of its bob and is given by .2 gLT π=

Express the period of pendulum A:

gLT A

A 2π=

Express the period of pendulum B:

gLT B

B 2π=

Divide the first of these equations by the second and solve for LA/LB:

2

B

A

B

A⎟⎟⎠

⎞⎜⎜⎝

⎛=

TT

LL

Substitute for TA and solve for LB to obtain: BB

2

B

BA 42 LL

TTL =⎟⎟

⎠

⎞⎜⎜⎝

⎛=

and correct. is )(c

Estimation and Approximation 21 •• Picture the Problem The Q factor for this system is related to the decay constant τ through TQ πττω 20 == and the amplitude of the child’s damped motion varies with

time according to .20

τteAA −= We can set the ratio of two displacements separated by

eight periods equal to 1/e to determine τ in terms of T.

Express Q as a function of τ : T

q πττω 20 == (1)

Oscillations

1053

The amplitude of the oscillations varies with time according to:

τ20

teAA −=

The amplitude after eight periods is:

( ) τ2808

TteAA +−=

Express and simplify the ratio A8/A: ( )τ

τ

τT

t

Tt

eeA

eAAA 4

20

2808 −

−

+−

==

Set this ratio equal to 1/e and solve for τ :

Tee T 414 =⇒= −− ττ

Substitute in equation (1) and evaluate Q:

( ) ππ 842==

TTQ

*22 •• Picture the Problem Assume that an average length for an arm is about 0.8 m, and that it can be treated as a uniform stick, pivoted at one end. We can use the expression for the period of a physical pendulum to derive an expression for the period of the swinging arm. When carrying a heavy briefcase, the mass is concentrated mostly at the end of the pivot (i.e., in the briefcase), so we can treat the arm-plus-briefcase as a simple pendulum. (a) Express the period of a uniform rod pivoted at one end: MgD

IT π2=

where I is the moment of inertia of the stick about an axis through one end, M is the mass of the stick, and D (= L/2) is the distance from the end of the stick to its center of mass.

Express the moment of inertia of the stick with respect to an axis through its end:

231 MLI =

Substitute the values for I and D to find T: ( ) g

LLMg

MLT3222

21

231

ππ ==


( )( ) s47.1

m/s81.93m8.022 2 == πT

(b) Express the period of a simple pendulum:

gL'T π2'=

where L′ is slightly longer than the arm

Chapter 14

1054

length due to the size of the briefcase.

Assuming L′ = 1 m, evaluate the period of the simple pendulum: s01.2

m/s81.9m12 2 == πT'

.reasonable seem estimates these, they walkas people ofn observatio From

Simple Harmonic Motion 23 • Picture the Problem The position of the particle is given by ( )δω += tAx cos where A

is the amplitude of the motion, ω is the angular frequency, and δ is a phase constant.

(a) Use the definition of ω to determine f:

Hz00.32

s62

1

===−

ππ

πωf

(b) Evaluate the reciprocal of the frequency:

s333.0Hz00.3

11===

fT

(c) Compare x = (7 cm) cos 6π t to

( )δω += tAx cos : cm00.7=A

(d) x = 0 when cosω t = 0:

20cos 1 πω == −t

Solve for t:

( ) s0833.0622

===ππ

ωπt

Differentiate x to find v(t): ( )[ ]

( ) t

tdtdv

ππ

π

6sincm/s42

6coscm7

−=

=

Evaluate v(0.0833 s):

( ) ( ) ( ) 0s0833.06sincm/s42s0833.0 <−= ππv

s. 0.0833 at direction negative in the moving is particle the0, Because =< tv

24 • Picture the Problem The initial position of the oscillating particle is related to the amplitude and phase constant of the motion by δcos0 Ax = where 0 ≤ δ < 2π.

Oscillations

1055

(a) For x0 = 0: 0cos =δ and

23,

20cos 1 ππδ == −

(b) For x0 = −A: δcosAA =−

and ( ) πδ =−= − 1cos 1

(c) For x0 = A: δcosAA =

and ( ) 01cos 1 == −δ

(d) When x = A/2: δcos

2AA

=

and

321cos 1 πδ =⎟⎠⎞

⎜⎝⎛= −

*25 • Picture the Problem The position of the particle as a function of time is given by ( )δω += tAx cos . Its velocity as a function of time is given by ( )δωω +−= tAv sin and its acceleration by ( )δωω +−= tAa cos2 . The initial position and velocity give us

two equations from which to determine the amplitude A and phase constantδ.

(a) Express the position, velocity, and acceleration of the particle as a function of t:

( )δω += tAx cos (1) ( )δωω +−= tAv sin (2) ( )δωω +−= tAa cos2 (3)

Find the angular frequency of the particle’s motion:

11 s19.4s3

42 −− ===ππω

T

Relate the initial position and velocity to the amplitude and phase constant:

δcos0 Ax =

and δω sin0 Av −=

Divide these equations to eliminate A:

δωδδω tan

cossin

0

0 −=−

=A

Axv

Chapter 14

1056

Solve for δ and substitute numerical values to obtain: 00tantan

0

1

0

01 =⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−= −−

ωωδ

xxv

Substitute in equation (1) to obtain: ( )

( ) ( )[ ]t

tx

1

1

s19.4coscm25

s3

4coscm25

−

−

=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=π

(b) Substitute in equation (2) to obtain:

( )

( ) ( )[ ]t

tv

1

11

s19.4sincm/s105

s3

4sins3

4cm25

−

−−

−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−=

ππ

(c) Substitute in equation (3) to obtain: ( )

( ) ( )[ ]t

ta

12

12

1

s19.4coscm/s439

s3

4coss3

4cm25

−

−−

−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−=

ππ

26 • Picture the Problem The maximum speed and maximum acceleration of the particle in are given by ωAv =max and .2

max ωAa = The particle’s position is given by ( )δω += tAx cos where A = 7 cm, ω = 6π s−1, and δ = 0, and its velocity is given by

( )δωω +−= tAv sin .

(a) Express vmax in terms of A and ω: ( )( )

m/s32.1cm/s42

s6cm7 1max

==

== −

π

πωAv

(b) Express amax in terms of A and ω: ( )( )

222

212max

m/s9.24cm/s252

s6cm7

==

== −

π

πωAa

(c) When x = 0: 0cos =tω

and

23,

20cos 1 ππω == −t

Evaluate v at :2πω =t ωπω AAv −=⎟

⎠⎞

⎜⎝⎛−=

2sin

i.e., the particle is moving to the left.

Oscillations

1057

Evaluate v at :2

3πω =t ωπω AAv =⎟⎠⎞

⎜⎝⎛−=

23sin

i.e., the particle is moving to the right.

Solve for t: ( ) s250.0

s623

23

1 === −ππ

ωπt

27 •• Picture the Problem The position of the particle as a function of time is given by

( )δω += tAx cos . Its velocity as a function of time is given by ( )δωω +−= tAv sin and its acceleration by ( )δωω +−= tAa cos2 . The initial position and velocity give us

two equations from which to determine the amplitude A and phase constant δ. (a) Express the position, velocity, and acceleration of the particle as functions of t:

( )δω += tAx cos (1) ( )δωω +−= tAv sin (2) ( )δωω +−= tAa cos2 (3)


11 s19.4s3

42 −− ===ππω

T

Relate the initial position and velocity to the amplitude and phase constant:

δcos0 Ax =

and δω sin0 Av −=

Divide these equations to eliminate A:

δωδδω tan

cossin

0

0 −=−

=A

Axv

Solve for δ:

⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

ωδ

0

01tanxv

Substitute numerical values and evaluate δ: ( )( )

rad445.0s24.19cm25

cm/s50tan 11

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−= −

−δ

Use either the x0 or v0 equation (x0 is used here) to find the amplitude: ( ) cm7.27

rad0.445coscm25

cos0 =

−==

δxA

Substitute in equation (1) to obtain: ( ) ( )[ ]445.0s19.4coscm7.27 1 −= − tx

Chapter 14

1058

(b) Substitute in equation (2) to obtain: ( )

( ) ( )[ ]445.0s19.4sincm/s116

445.0s3

4sin

s3

4cm7.27

1

1

1

−−=

⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛×

⎟⎠⎞

⎜⎝⎛−=

−

−

−

t

t

v

π

π

(c) Substitute in equation (3) to obtain:

( )

( ) ( )[ ]445.0s19.4coscm/s486

445.0s3

4coss3

4cm7.27

12

12

1

−−=

⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−=

−

−−

t

ta ππ

28 •• Picture the Problem The position of the particle as a function of time is given by ( )δω += tAx cos . We’re given the amplitude A of the motion and can use the initial

position of the particle to determine the phase constant δ. Once we’ve determined these quantities, we can express the distance traveled ∆x during any interval of time.

Express the position of the particle as a function of t:

( ) ( )δω += tx coscm12 (1)


1s4s8

22 −===πππω

T

Relate the initial position of the particle to the amplitude and phase constant:

δcos0 Ax =

Solve for δ: 2

0coscos 101 πδ === −−

AAx

Substitute in equation (1) to obtain: ( ) ⎥

⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛= −

2s

4coscm12 1 ππ tx

Oscillations

1059

Express the distance the particle travels in terms of tf and ti:

( )

( )

( )

⎭⎬⎫⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛=

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛=∆

−

−

−

−

2s

4cos

2s

4coscm12

2s

4coscm12

2s

4coscm12

i1

f1

i1

f1

ππ

ππ

ππ

ππ

t

t

t

tx

(a) Evaluate ∆x for tf = 2 s, ti = 1 s:

( ) ( )

( )

( ) cm0.12

01cm12

20s

4cos

2s2s

4coscm12

1

1

=

−−=⎭⎬⎫⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛=∆

−

−

ππ

ππx

(b) Evaluate ∆x for tf = 4 s, ti = 2 s:

( ) ( )

( )

( ) cm0.12

10cm12

2s2s

4cos

2s4s

4coscm12

1

1

=

−=⎭⎬⎫⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛=∆

−

−

ππ

ππx

(c) Evaluate ∆x for tf = 1 s, ti = 0:

( ) ( )

( )

( ) cm49.8

07071.0cm12

20s

4cos

2s1s

4coscm12

1

1

=

−−=⎭⎬⎫⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛=∆

−

−

ππ

ππx

Chapter 14

1060

(d) Evaluate ∆x for tf = 2 s, ti = 1 s: ( ) ( )

( )

( ) cm51.3

7071.01cm12

2s1s

4cos

2s2s

4coscm12

1

1

=

+−=⎭⎬⎫⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛=∆

−

−

ππ

ππx

29 •• Picture the Problem The position of the particle as a function of time is given by ( ) ( )δω += tx coscm10 . We can determine the angular frequency ω from the period

of the motion and the phase constant δ from the initial position and velocity. Once we’ve determined these quantities, we can express the distance traveled ∆x during any interval of time. Express the position of the particle as a function of t:

( ) ( )δω += tx coscm10 (1)


1s4s8

22 −===πππω

T

Find the phase constant of the motion: 00tantan

0

1

0

01 =⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−= −−

ωωδ

xxv

Substitute in equation (1) to obtain: ( ) ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛= − tx 1s

4coscm10 π

Oscillations

1061

(a) A graph of ( ) ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛= − tx 1s

4coscm10 π

follows:

-10

-8

-6

-4

-2

0

2

4

6

8

10

0 2 4 6 8

t (s)

x (c

m)

(b) Express the distance the particle travels in terms of tf and ti:

( ) ( )

( )⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=∆

−−

−−

i1

f1

i1

f1

s4

coss4

coscm10

s4

coscm10s4

coscm10

tt

ttx

ππ

ππ

(2)

Substitute numerical values in equation (2) and evaluate ∆x in each of the given time intervals to obtain:

tf ti ∆x (s) (s) (cm) 1 0 93.2

2 1 07.7

3 2 07.7

4 3 93.2

*30 •• Picture the Problem We can use the expression for the maximum acceleration of an oscillator to relate the 10g military specification to the compliance frequency. Express the maximum acceleration of an oscillator:

2max ωAa =

Chapter 14

1062

Express the relationship between the angular frequency and the frequency of the vibrations:

fπω 2=


22max 4 Afa π=

Solve for f:

Aaf max

21π

=

Substitute numerical values and evaluate f: Hz9.12

m101.5m/s98.1

21

2

2

=×

= −πf

31 •• Picture the Problem The maximum speed and acceleration of the particle are given by

ωAv =max and 2max ωAa = . The velocity and acceleration of the particle are given by

tAv ωω sin−= and .cos2 tAa ωω−=

(a) Find vmax from A and ω: ( )( )m/s85.7

sm5.2 1max

=

== −πωAv

Find amax from A and ω: ( )( )

2

212max

m/s7.24

sm5.2

=

== −πωAa

(b) Use the equation for the position of the particle to relate its position at x = 1.5 m to the time t′to reach this position:

( ) t'πcosm2.5m5.1 =

Solve for π t′: rad9273.06.0cos 1 == −t'π

Evaluate v when π t = π t′: ( )( ) ( )

m/s28.6

rad9273.0sinsm5.2 1

−=

−= −πv

where the minus sign indicates that the particle is moving in the negative direction.

Evaluate a when πt = πt′: ( )( ) ( )2

21

m/s8.14

rad9273.0cossm5.2

−=

−= −πa

Oscillations

1063

where the minus sign indicates that the particle’s acceleration is in the negative direction.

*32 •• Picture the Problem We can use the formula for the cosine of the sum of two angles to write x = A0 cos(ωt + δ) in the desired form. We can then evaluate x and dx/dt at t = 0 to relate Ac and As to the initial position and velocity of a particle undergoing simple harmonic motion. (a) Apply the trigonometric identity

( ) δωδωδω sinsincoscoscos ttt −=+ to obtain:

( ) []

tAtA

tAtA

ttAtAx

ωω

ωδωδδω

δωδω

cossin

coscossinsin

sinsincoscoscos

cs

0

0

00

+=

+−=−

=+=

provided δsin0s AA −= and δcos0c AA =

(b) At t = 0:

c0 cos)0( AAx == δ

Evaluate dx/dt: [ ]

tAtA

tAtAdtd

dtdxv

ωωωω

ωω

sincos

cossin

cs

cs

−=

+==

Evaluate v(0) to obtain: δωω sin)0( 0AAv s −==

Simple Harmonic Motion and Circular Motion 33 • Picture the Problem We can find the period of the motion from the time required for the particle to travel completely around the circle. The frequency of the motion is the reciprocal of its period and the x-component of the particle’s position is given by ( )δω += tAx cos .

(b) Use the definition of speed to find the period of the motion:

( ) s14.3m/s8.0

m4.022===

ππv

rT

(a) Because the frequency and the period are reciprocals of each other:

Hz318.0s14.3

11===

Tf

(c) Express the x component of the ( )δω += tAx cos

Chapter 14

1064

position of the particle:

Assuming that the particle is on the positive x axis at time t = 0:

δcosAA = ⇒ 01cos 1 == −δ

Substitute for A, ω, and δ to obtain: ( )( ) ( )[ ]t

ftAx1s2coscm40

2cos−=

= π

*34 • Picture the Problem We can find the period of the motion from the time required for the particle to travel completely around the circle. The angular frequency of the motion is 2π times the reciprocal of its period and the x-component of the particle’s position is given by ( )δω += tAx cos .

(a) Use the definition of speed to express and evaluate the speed of the particle:

( ) cm/s4.31s3cm1522

===ππ

Trv

(b) Express the angular velocity of the particle:

rad/s322 ππω ==

T

(c) Express the x component of the position of the particle:

( )δω += tAx cos

Assuming that the particle is on the positive x axis at time t = 0:

δcosAA = ⇒ 01cos 1 == −δ

Substitute to obtain: ( ) tx ⎟⎠⎞

⎜⎝⎛= −1s

32coscm15 π

Energy in Simple Harmonic Motion 35 • Picture the Problem The total energy of the object is given by ,2

21

tot kAE = where A is

the amplitude of the object’s motion. Express the total energy of the system:

221

tot kAE =

Substitute numerical values and evaluate Etot:

( )( ) J22.5m0.1kN/m4.5 221

tot ==E

Oscillations

1065

36 • Picture the Problem The total energy of an oscillating object can be expressed in terms of its kinetic energy as it passes through its equilibrium position: .2

max21

tot mvE = Its

maximum speed, in turn, can be expressed in terms of its angular frequency and the amplitude of its motion. Express the total energy of the object in terms of its maximum kinetic energy:

2max2

1 mvE =

Express vmax: AfAv πω 2max ==

Substitute to obtain: ( ) 2222

21 22 fmAAfmE ππ ==

Substitute numerical values and evaluate E:

( )( ) ( )J41.3

s4.2m1.0kg32 2122

=

= −πE

37 • Picture the Problem The total mechanical energy of the oscillating object can be expressed in terms of its kinetic energy as it passes through its equilibrium position:

2max2

1tot mvE = . Its total energy is also given by .2

21

tot kAE = We can equate these

expressions to obtain an expression for A. (a) Express the total mechanical energy of the object in terms of its maximum kinetic energy:

2max2

1 mvE =


( )( ) J0.368m/s0.7kg1.5 221 ==E

(b) Express the total energy of the object in terms of the amplitude of its motion:

221

tot kAE =

Solve for A:

kEA tot2

=


( ) cm84.3N/m500

J368.02==A

Chapter 14

1066

38 • Picture the Problem The total energy of the oscillating object can be expressed in terms of its kinetic energy as it passes through its equilibrium position: .2

max21

tot mvE = Its total

energy is also given by .221

tot kAE = We can solve the latter equation to find A and solve

the former equation for vmax. (a) Express the total energy of the object as a function of the amplitude of its motion:

221

tot kAE =

Solve for A:

kEA tot2

=


( ) cm00.3N/m2000

J9.02==A

(b) Express the total energy of the object in terms of its maximum speed:

2max2

1tot mvE =

Solve for vmax:

mEv tot

max2

=

Substitute numerical values and evaluate vmax:

( ) m/s0.775kg3

J0.92max ==v

39 • Picture the Problem The total energy of the object is given by .2

21

tot kAE = We can

solve this equation for the force constant k and substitute the numerical data to determine its value. Express the total energy of the oscillator as a function of the amplitude of its motion:

221

tot kAE =

Solve for k: 2tot2

AEk =


( )( )

kN/m1.38m0.045J1.42

2 ==k

Oscillations

1067

*40 •• Picture the Problem The total energy of the object is given, in terms of its maximum kinetic energy by .2

max21

tot mvE = We can express vmax in terms of A and ω and, in turn,

express ω in terms of amax to obtain an expression for Etot in terms of amax. Express the total energy of the object in terms of its maximum kinetic energy:

2max2

1tot mvE =

Relate the maximum speed of the object to its angular frequency:

ωAv =max


( ) 22212

21

tot ωω mAAmE ==

Relate the maximum acceleration of the object to its angular frequency:

2max ωAa =

or

Aamax2 =ω


max21max2

21

tot mAaA

amAE ==

Substitute numerical values and evaluate Etot:

( )( )( )J0.420

m/s3.50m0.08kg3 221

tot

=

=E

Springs 41 •

Picture the Problem The frequency of the object’s motion is given by .21 mkfπ

=

Its period is the reciprocal of its frequency. The maximum velocity and acceleration of an object executing simple harmonic motion are ωAv =max and ,2

max ωAa = respectively.

(a) The frequency of the motion is given by: m

kfπ21

=


kg2.4kN/m4.5

21

==π

f

Chapter 14

1068

(b) The period of the motion to is the reciprocal of its frequency:

s145.0s89.6

111 === −f

T

(c) Because the object is released from rest after the spring to which it is attached is stretched 10 cm:

m100.0=A

(d) Express the object’s maximum speed:

fAAv πω 2max ==


( )( ) m/s33.4m1.0s89.62 1max == −πv

(e) Express the object’s maximum acceleration:

maxmax2

max 2 fvvAa πωω ===

Substitute numerical values and evaluate amax:

( )( )2

1max

m/s187

m/s33.4s89.62

=

= −πa

(f) The object first reaches its equilibrium when:

( ) ms3.36s145.041

41 === Tt

Because the resultant force acting on the object as it passes through its equilibrium point is zero, the acceleration of the object is:

0=a

42 •

Picture the Problem The frequency of the object’s motion is given by .21 mkfπ

=

Its period is the reciprocal of its frequency. The maximum velocity and acceleration of an object executing simple harmonic motion are ωAv =max and ,2


(a) The frequency of the motion is given by: m

kfπ21

=


kg5N/m700

21

==π

f

Oscillations

1069

(b) The period of the motion is the reciprocal of its frequency:

s531.0s88.1

111 === −f

T

(c) Because the object is released from rest after the spring to which it is attached is stretched 8 cm:

m0800.0=A

(d) Express the object’s maximum speed:

fAAv πω 2max ==


( )( ) m/s945.0m08.0s88.12 1max == −πv

(e) Express the object’s maximum acceleration:

maxmax2

max 2 fvvAa πωω ===


( )( )2

1max

m/s2.11

m/s945.0s88.12

=

= −πa

(f) The object first reaches its equilibrium when:

( ) s.1330s531.041

41 === Tt

Because the resultant force acting on the object as it passes through its equilibrium point is zero, the acceleration of the object is:

0=a

43 • Picture the Problem The angular frequency, in terms of the force constant of the spring and the mass of the oscillating object, is given by .2 mk=ω The period of the motion is

the reciprocal of its frequency. The maximum velocity and acceleration of an object executing simple harmonic motion are ωAv =max and ,2


(a) Relate the angular frequency of the motion to the force constant of the spring:

mk

=2ω

or mfmk 222 4πω ==


( ) ( ) N/m682kg3s4.24 212 == −πk

Chapter 14

1070

(b) Relate the period of the motion to its frequency:

s417.0s4.2

111 === −f

T

(c) Express the maximum speed of the object:

fAAv πω 2max ==


( )( ) m/s51.1m1.0s4.22 1max == −πv

(d) Express the maximum acceleration of the object:

AfAa 222max 4πω ==


( ) ( ) 2212max m/s7.22m0.1s4.24 == −πa

*44 • Picture the Problem We can find the frequency of vibration of the car-and-passenger

system using ,21

Mkf

π= where M is the total mass of the system. The spring

constant can be determined from the compressing force and the amount of compression. Express the frequency of the car-and-passenger system: M

kfπ21

=

Express the spring constant:

xmg

xFk

∆=

∆=

where m is the person’s mass.


xMmgf∆

=π21

Substitute numerical values and evaluate f:

( )( )( )( )

Hz601.0

m102.35kg2485m/s9.81kg85

21

2

2

=

×= −π

f

45 • Picture the Problem We can relate the force constant k to the maximum acceleration by eliminating ω2 between mk=2ω and .2

max ωAa = We can also express the frequency f

Oscillations

1071

of the motion by substituting mamax/A for k in .21

mkf

π=

(a) Relate the angular frequency of the motion to the force constant and the mass of the oscillator:

mk

=2ω or mk 2ω=

Relate the object’s maximum acceleration to its angular frequency and amplitude and solve for the square of the angular frequency:

2max ωAa =

or

Aamax2 =ω (1)


Amak max=


( )( ) kN/m08.3m103.8

m/s26kg4.52

2

=×

= −k

(b) Replace ω in equation (1) by 2πf and solve for f to obtain: A

af max

21π

=


m103.8m/s26

21

2

2

=×

= −πf

(c) The period of the motion is the reciprocal of its frequency:

s240.0s16.4

111 === −f

T

46 • Picture the Problem We can find the frequency of the motion from its maximum speed and the relationship between frequency and angular frequency. The mass of the object can be found by eliminating ω between mk=2ω and .max ωAv =

(b) Express the object’s maximum speed as a function of the frequency of its motion:

fAAv πω 2max == (1)

Solve for f: A

vfπ2max=

Substitute numerical values and evaluate f: ( ) Hz04.6

m108.52m/s2.2

2 =×

= −πf

Chapter 14

1072

(a) Relate the square of the angular frequency of the motion to the force constant and the mass of the object:

mk

=2ω ⇒ 2ωkm = (2)

Eliminate ω between equations (1) and (2) to obtain: 2

max

2

vkAm =


( )( )( )

kg25.1

m/s2.2m105.8N/m101.8

2

223

=

××=

−

m

(c) The period of the motion is the reciprocal of its frequency:

s166.0s04.6

111 === −f

T

47 •• Picture the Problem The maximum speed of the block is given by ωAv =max and the

angular frequency of the motion is rad/s48.5== mkω . We’ll assume that the

position of the block is given by tAx ωcos= and solve for ωt for x = 4 cm and x = 0. We can use these values for ωt to find the time for the block to travel from x = 4 cm to its equilibrium position. (a) Express the maximum speed of the block as a function of the system’s angular frequency:

ωAv =max


( )( )m/s438.0

rad/s48.5m08.0max

=

=v

(b) Assuming that t,Ax ωcos=

evaluate ωt for x = 4 cm = A/2: 321coscos

21 πωω ==⇒= −ttAA

Evaluate v for :3πω =t ( )

( ) m/s379.023m/s438.0

3sinm/s438.0sinmax

==

==πωtvv

Express a as a function of vmax and ω:

tvtAa ωωωω coscos max2 ==

Oscillations

1073


( )( )2m/s20.1

3cosrad/s48.5m/s438.0

=

=πa

(c) Evaluate ωt for x = 0:

20coscos0 1 πωω ==⇒= −ttA

Let ∆t = time to go from 3πω =t to

2πω =t . Then: 632πππω =−=∆t

Solve for and evaluate ∆t:

( ) ms5.95rad/s48.566

===∆π

ωπt

*48 •• Picture the Problem Choose a coordinate system in which upward is the positive y direction. We can find the mass of the object using .2ωkm = We can apply a condition

for translational equilibrium to the object when it is at its equilibrium position to determine the amount the spring has stretched from its natural length. Finally, we can use the initial conditions to determine A and δ and express x(t) and then differentiate this expression to obtain v(t) and a(t). (a) Express the angular frequency of the system in terms of the mass of the object fastened to the vertical spring and solve for the mass of the object:

22

ωω km

mk

=⇒=

Express ω2 in terms of f: 222 4 fπω =


224 fkm

π=

Substitute numerical values and evaluate m: ( ) kg51.1

s5.54N/m1800

212==

−πm

(b) Letting ∆x represent the amount the spring is stretched from its natural length when the object is in equilibrium, apply ∑ = 0yF to the

object when it is in equilibrium:

0=−∆ mgxk

Chapter 14

1074

Solve for ∆x: k

mgx =∆

Substitute numerical values and evaluate ∆x:

( )( ) mm23.8N/m 1800

m/s81.9kg51.1 2

==∆x

(c) Express the position of the object as a function of time:

( )δω += tAx cos

Use the initial conditions (x0 = −2.5 cm and v0 = 0) to find δ: π

ωδ ==⎟⎟

⎠

⎞⎜⎜⎝

⎛−= −− 0tantan 1

0

01

xv

Evaluate ω:

rad/s34.5kg1.51N/m1800

===mkω

Substitute to obtain: ( ) ( )[ ]( ) ( )[ ]t

tx

rad/s5.34coscm5.2

rad/s5.34coscm5.2

−=

+= π

Differentiate x(t) to obtain v: ( ) ( )[ ]tv rad/s5.34sincm/s4.86=

Differentiate v(t) to obtain a: ( ) ( )[ ]ta rad/s5.34cosm/s8.29 2=

49 •• Picture the Problem Let the system include the object and the spring. Then, the net external force acting on the system is zero. Choose Ei = 0 and apply the conservation of mechanical energy to the system. Express the period of the motion in terms of its angular frequency:

ωπ2

=T (1)

Apply conservation of energy to the system:

fi EE = or springg0 UU +=

Substitute for Ug and Uspring: ( )2210 xkxmg ∆+∆−=

Solve for ω2 = k/m:

xg

mk

∆==

22ω

Substitute numerical values and evaluate ω2:

( ) 22

22 rad/s574

m103.42m/s9.812

=×

= −ω

Oscillations

1075

Substitute in equation (1) to obtain: s0.262rad/s574

2π2==T

50 •• Picture the Problem Let the system include the object and the spring. Then the net external force acting on the system is zero. Because the net force acting on the object when it is at its equilibrium position is zero, we can apply a condition for translational equilibrium to determine the distance from the starting point to the equilibrium position. Letting Ei = 0, we can apply conservation of energy to the system to determine how far down the object moves before coming momentarily to rest. We can find the period of the motion and the maximum speed of the object from kmT π2= and .max mkAv =

(a) Apply ∑ = 0yF to the object

when it is at the equilibrium position:

00 =−mgky

Solve for y0: k

mgy =0

Substitute numerical values and evaluate y0:

( )( ) cm3.92N/m250

m/s9.81kg1 2

0 ==y

(b) Apply conservation of energy to the system:

fi EE =

or springg0 UU +=

Substitute for Ug and Uspring: 2

f21

f0 kymgy +−=

Solve for yf:

kmgy 2

f =

Substitute numerical values and evaluate yf:

( )( ) cm7.85N/m250

m/s9.81kg12 2

f ==y

(c) Express the period T of the motion in terms of the mass of the object and the spring constant:

kmT π2=

Chapter 14

1076

Substitute numerical values and evaluate T: s0.397

N/m250kg1π2 ==T

(d) The object will be moving with its maximum speed when it reaches its equilibrium position:

mkAAv == ωmax


( )

cm/s62.0

kg1N/m250cm3.92max

=

=v

(e) The time required for the object to reach equilibrium is one-fourth of its period:

( ) ms99.3s0.39741

41 === Tt

51 •• Picture the Problem The stunt woman’s kinetic energy, after 2 s of flight, is

.2s22

1s2 mvK = We can evaluate this quantity as soon as we know how fast she is moving

after two seconds. Because her motion is oscillatory, her velocity as a function of time is ( ) ( ).sin δωω +−= tAtv We can find the amplitude of her motion from her distance of

fall and the angular frequency of her motion by applying conservation of energy to her fall to the ground. Express the kinetic energy of the stunt woman when she has fallen for 2 s:

2s22

1s2 mvK = (1)

Express her velocity as a function of time:

( ) ( )δωω +−= tAtv sin

where δ = 0 (she starts from rest with positive displacement) and

( ) m96m19221 ==A ( ) ( ) ( )ttv ωω sinm96−=∴ (2)

Letting Ei = 0, use conservation of energy to find the force constant of the elastic band:

elasticg0 UU +=

or 00 2

21 =+−= khmgh

Solve for k: hmgk 2

=

Oscillations

1077


( )( ) N/m6.13m192

m/s9.81kg602 2

==k

Express the angular frequency of her motion: m

k=ω


kg60N/m6.13

==ω

Substitute in equation (2) to obtain: ( ) ( )( )

( )[ ]( ) ( )[ ]t

ttv

rad/s320.0sinm/s7.30rad/s320.0sin

rad/s320.0m96

=×−=

Evaluate v(2 s): ( ) ( ) ( )( )[ ]

m/s3.18s2rad/s320.0sinm/s7.30s2

==v

Substitute in equation (1) and evaluate K(2 s):

( ) ( )( ) kJ10.1m/s18.3kg60s2 221 ==K

*52 •• Picture the Problem The diagram shows the stretched bungie cords supporting the suitcase under equilibrium conditions. We

can use Mkf eff

21π

= to express the

frequency of the suitcase in terms of the effective ″spring″ constant keff and apply a condition for translational equilibrium to the suitcase to find keff. Express the frequency of the suitcase oscillator: M

kf eff

21π

=

Apply 0=∑ yF to the suitcase to

obtain:

0=−+ Mgkxkx

or 02 =− Mgkx

or 0eff =− Mgxk

where keff = 2k

Chapter 14

1078

Solve for keff to obtain: x

Mgk =eff


xgf

π21

=


m05.0m/s81.9

21 2

==π

f

53 •• Picture the Problem The frequency of the motion of the stone and block depends on the force constant of the spring and the mass of the stone plus block. The force constant can be determined from the equilibrium of the system when the spring is stretched additionally by the addition of the stone to the mass. When the block is at the point of maximum upward displacement, it is momentarily at rest and the net force acting on it is its weight. (a) Express the frequency of the motion in terms of k and m:

tot21

mkf

π=

where mtot is the total mass suspended from the spring.

Apply ∑ = 0yF to the stone when

it is at its equilibrium position:

0=−∆ mgyk

Solve for k: y

mgk∆

=


( )( ) N/m5.89m0.05

m/s9.81kg0.03 2

==k

Substitute and evaluate f:

Hz997.0kg0.15

N/m5.8921

==π

f

(b) The time to travel from its lowest point to its highest point is one-half its period:

( ) s0.502s0.9972

121

121 ==== −fTt

Oscillations

1079

(c) When the stone is at a point of maximum upward displacement:

( )( )N0.294

m/s9.81kg03.0 2net

=

== mgF

54 •• Picture the Problem We can use the maximum acceleration of the oscillator

2max ωAa = to express amax in terms of A, k, and m. k can be determined from the

equilibrium of the system when the spring is stretched additionally by the addition of the stone to the mass. If the stone is to remain in contact with the block, the block’s maximum downward acceleration must not exceed g. Express the maximum acceleration in terms of the angular frequency and amplitude of the motion:

2max ωAa =

Relate ω2 to the force constant and the mass of the stone: m

k=2ω


mkAa =max

Apply ∑ = 0yF to the stone when

it is at its equilibrium position:

0=−∆ mgyk

Solve for k: y

mgk∆

=


( )( ) N/m5.89m0.05

m/s9.81kg0.03 2

==k

Substitute numerical values to express amax in terms of A:

( )AAa 2max s3.39

kg0.15N/m5.89 −==

Set amax = g and solve for Amax:

2max s3.39 −=gA

Substitute for g and evaluate Amax: cm0.25

s39.3m/s9.81

2

2

max == −A

Chapter 14

1080

55 •• Picture the Problem The maximum height above the floor to which the object rises is the sum of its initial distance from the floor and the amplitude of its motion. We can find the amplitude of its motion by relating it to the object’s maximum speed. Because the object initially travels downward, it will be three-fourths of the way through its cycle when it first reaches its maximum height. We can find the minimum initial speed the object would need to be given in order for the spring to become uncompressed by applying conservation of energy. (a) Relate h, the maximum height above the floor to which the object rises, to the amplitude of its motion:

h = A + 5.0 cm (1)

Relate the maximum speed of the object to the angular frequency and amplitude of its motion and solve for the amplitude:

ωAv =max

or

kmvA max= (2)

Using its definition, express and evaluate the force constant of the spring:

( )( ) N/m654m0.03

m/s9.81kg2 2

==∆

=y

mgk

Substitute numerical values in equation (2) and evaluate A: cm1.66

N/m654kg2m/s3.0 ==A

Substitute in equation (1) to obtain: cm6.66cm5.00cm66.1 =+=h

(b) Express the time required for the object to reach its maximum height the first time:

Tt 43=

Express the period of the motion:

kmT π2=


N/m654kg22 == πT

Substitute to obtain: ( ) s261.0s347.04

3 ==t

Oscillations

1081

(c) Because h < 8.0 cm: ed.uncompressnever is spring the

Using conservation of energy and letting Ug be zero 5 cm above the floor, relate the height to which the object rises, ∆y, to its initial kinetic energy:

0sg =∆+∆+∆ UUK

or, because Kf = Ui = 0, ( )

( ) 02i2

1

2212

i21

=−−

∆+∆−

yLk

ykymgmv

Because :iyLy −=∆

( ) ( ) 02

212

212

i21 =∆−∆+∆− ykykymgmv

and 02

i21 =∆− ymgmv

Solve for and evaluate vi for ∆y = 3 cm:

( )( )m/s0.767

cm3m/s9.8122 2i

=

=∆= ygv

i.e., the minimum initial velocity that must be given to the object for the spring to be uncompressed at some time is

m/s767.0

*56 •• Picture the Problem We can relate the elongation of the cable to the load on it using the definition of Young’s modulus and use the expression for the frequency of a spring and mass oscillator to find the oscillation frequency of the engine block at the end of the wire. (a) Using the definition of Young’s modulus, relate the elongation of the cable to the applied stress:

ll∆==

AFYstrainstress

Solve for ∆l: AY

MgAYF ll

l ==∆

Substitute numerical values and evaluate ∆l:

( )( )( )( )( )

mm04.1

GN/m150cm5.1m5.2m/s81.9kg950

22

2

=

=∆l

(b) Express the oscillation frequency of the wire-engine block system:

Mkf eff

21π

=

Chapter 14

1082

Express the effective ″spring″ constant of the cable: ll ∆

=∆

=MgFkeff


l∆=

gfπ21


mm04.1m/s81.9

21 2

==π

f

Energy of an Object on a Vertical Spring 57 •• Picture the Problem Let the origin of our coordinate system be at y0, where y0 is the equilibrium position of the object and let Ug = 0 at this location. Because Fnet = 0 at equilibrium, the extension of the spring is then y0 = mg/k, and the potential energy stored in the spring is 2

021

s kyU = . A further extension of the spring by an amount y increases Us

to ( ) .202

12212

021

02

212

021 kymgykykykyykyyyk ++=++=+ Consequently, if we set

U = Ug + Us = 0, a further extension of the spring by y increases Us by ½ky2 + mgy while decreasing Ug by mgy. Therefore, if U = 0 at the equilibrium position, the change in U is given by ( ) ,' 2

21 yk where y′ = y − y0.

(a) Express the total energy of the system:

221 kAE =


( )( ) J0.270m0.03N/m600 221 ==E

(b) Express and evaluate Ug when the object is at its maximum downward displacement:

( )( )( )J0.736

m0.03m/s9.81kg2.5 2

g

−=

−=

−= mgAU

(c) When the object is at its maximum downward displacement: ( )( )

( )( )( )J1.01

m0.03m/s9.81kg2.5

m0.03N/m6002

221

221

s

=

+

=

+= mgAkAU

Oscillations

1083

(d) The object has its maximum kinetic energy when it is passing through its equilibrium position:

( )( )J0.270

m0.03N/m600 2212

21

max

=

== kAK

58 •• Picture the Problem Let the origin of our coordinate system be at y0, where y0 is the equilibrium position of the object and let Ug = 0 at this location. Because Fnet = 0 at equilibrium, the extension of the spring is then y0 = mg/k, and the potential energy stored in the spring is .2

021

s kyU = A further extension of the spring by an amount y increases Us

to ( ) .202

12212

021

02

212




(a) Express the total energy of the system:

221 kAE =

Letting ∆y represent the amount the spring is stretched from its natural length by the 1.5-kg object, apply

∑ = yy maF to the object when it is

in its equilibrium position:

0=−∆ mgyk

Solve for k: y

mgk∆

=

Substitute for k to obtain:

ymgAE∆

=2

2


( )( )( )( )

J127.0

m0.0282m022.0m/s9.81kg1.5 22

=

=E

(b) Express Ug when the object is at its maximum downward displacement:

mgAU −=g

Substitute numerical values and evaluate Ug:

( )( )( )J0.324

m0.022m/s9.81kg1.5 2g

−=

−=U

Chapter 14

1084

(c) When the object is at its maximum downward displacement:

mgAkAU += 221

s

Substitute numerical values and evaluate Us:

( )( )( )( )( )

J451.0

m0.022m/s9.81kg1.5

m0.022N/m2652

221

s

=

+

=U

(d) The object has its maximum kinetic energy when it is passing through its equilibrium position:

( )( )J0.127

m0.022N/m265 221

221

max

=

=

= kAK

*59 •• Picture the Problem We can find the amplitude of the motion by relating it to the maximum speed of the object. Let the origin of our coordinate system be at y0, where y0 is the equilibrium position of the object and let Ug = 0 at this location. Because Fnet = 0 at equilibrium, the extension of the spring is then y0 = mg/k, and the potential energy stored in the spring is .2

021

s kyU = A further extension of the spring by an amount y increases Us

to ( ) .202

12212

021

02

212




(a) Relate the maximum speed of the object to the amplitude of its motion:

ωAv =max

Solve for A: k

mvvA maxmax ==ω

Substitute numerical values and evaluate A: ( ) cm1.90

N/m300kg1.2m/s0.3 ==A

(b) Express the energy of the object at maximum displacement:

221 kAE =


( )( ) J0.0542m0.019N/m300 221 ==E

Oscillations

1085

(c) At maximum displacement from equilibrium:

mgAU −=g

Substitute numerical values and evaluate Ug:

( )( )( )J0.224

m0.019m/s9.81kg1.2 2g

−=

−=U

(d) Express the potential energy in the spring when the object is at its maximum downward displacement:

mgAkAU += 221

s

Substitute numerical values and evaluate Us:

( )( )( )( )( )

J278.0

m019.0m/s81.9kg2.1

m0.019N/m3002

221

s

=

+

=U

Simple Pendulums 60 • Picture the Problem We can determine the required length of the pendulum from the expression for the period of a simple pendulum. Express the period of a simple pendulum:

gLT π2=

Solve for L:

2

2

4πgTL =

Substitute numerical values and evaluate L: ( ) ( ) m21.6

4m/s9.81s5

2

22

==π

L

61 • Picture the Problem We can find the period of the pendulum from moon2 gLT π= where gg 6

1moon = and L = 6.21 m.

Express the period of a simple pendulum:

moon

2g

LT π=

Substitute numerical values and evaluate T: ( ) s2.12

m/s9.81m6.212 2

61

== πT

Chapter 14

1086

62 • Picture the Problem We can find the value of g at the location of the pendulum by solving the equation gLT π2= for g and evaluating it for the given length and

period. Express the period of a simple pendulum: g

LT π2=

Solve for g:

2

24T

Lg π=


( )( )

22

2

m/s79.9s68.1

m7.04==

πg

*63 • Picture the Problem We can use gLT π2= to find the period of this pendulum.

Express the period of a simple pendulum: g

LT π2=


m/s9.81m432 2 == πT

64 •• Picture the Problem The figure shows the simple pendulum at maximum angular displacement φ0. The total energy of the simple pendulum is equal to its initial gravitational potential energy. We can apply the definition of gravitational potential energy and use the small-angle approximation to show that .2

021 φmgLE ≈

Express the total energy of the simple pendulum at maximum displacement: [ ]0

ntdisplacememax

cos1 φ−=

==

mgL

mghUE

For φ << 1: 2

211cos φφ −≈

Oscillations

1087

Substitute and simplify to obtain: ( )[ ] 202

1202

111 φφ mgLmgLE =−−=

65 •• Picture the Problem Because the cart is accelerating down the incline, the period of the simple pendulum will be given by

eff2 gLT π= where geff is less than g

by the acceleration of the cart. We can apply Newton’s 2nd law to the cart to find its acceleration down the incline and then subtract this acceleration from g to find geff. Express the period of a simple pendulum in terms of its length and the effective value of the acceleration of gravity:

eff

2gLT π=

Relate geff to the acceleration of the cart:

agg −=eff

Apply xx maF =∑ to the cart and

solve for its acceleration:

mamg =θsin

and θsinga =


( )θπ

θππ

sin12

sin22

−=

−=

−=

gL

ggL

agLT

Chapter 14

1088

66 •• Picture the Problem The figure shows the simple pendulum at maximum angular displacement φ0. We can express the angular position of the pendulum’s bob in terms of its initial angular position and time and differentiate this expression to find the maximum speed of the bob. We can use conservation of energy to find an exact value for vmax and the approximation

2211cos φφ −≈ to show that this value

reduces to the former value for small φ. (a) Relate the speed of the pendulum’s bob to its angular speed:

dtdLv φ

= (1)

Express the angular position of the pendulum as a function of time:

tωφφ cos0=

Differentiate this expression to express the angular speed of the pendulum:

tdtd ωωφφ sin0−=

Substitute in equation (1) to obtain: tvtLv ωωωφ sinsin max0 −=−=

Simplify vmax to obtain:

gLLgLv 00max φφ ==

(b) Use conservation of energy to relate the potential energy of the pendulum at point 1 to its kinetic energy at point 2:

0=∆+∆ UK or, because K1 = U2 = 0,

012 =−UK

Substitute for K2 and U1: 0222

1 =−mghmv

Express h in terms of L and φ0:

( )0cos1 φ−= Lh

Substitute for h and solve for v2 = vmax to obtain:

( )0max cos12 φ−= gLv (2)

Oscillations

1089

(c) For φ0 << 1: 202

10cos1 φφ ≈−

Substitute in equation (2) to obtain: ( ) gLgLv 0

202

1max 2 φφ ==

in agreement with our result in part (a).

(d) Express the difference in the results from (a) and (b):

bmax,amax, vvv −=∆ (3)

Using φ0 = 0.20 rad and L = 1 m, evaluate the result in (b):

( )( )( )m/s0.6254

2.0cos1m1m/s81.92 2bmax,

=

−=v

Using φ0 = 0.20 rad and L = 1 m, evaluate the result in part (a):

( ) ( )( )m/s6264.0

m1m/s9.81rad20.0 2a.max

=

=v


mm/s1.00m/s0.001

m/s0.6254m/s6264.0

==

−=∆v

Physical Pendulums 67 • Picture the Problem The period of this physical pendulum is given by

MgDIT π2= where I is the moment of inertia of the thin disk with respect to an

axis through its pivot point. We can use the parallel-axis theorem to express I in terms of the moment of inertia of the disk with respect to its center of mass and the distance from its center of mass to its pivot point. Express the period of physical pendulum: MgD

IT π2=

Using the parallel-axis theorem, find the moment of inertia of the thin disk about an axis through the pivot point:

223

22212

cm

MR

MRMRMRII

=

+=+=


gR

MgRMRT

2322

223

ππ ==


( )( ) s10.1

m/s9.812m0.232 2 == πT

Chapter 14

1090

68 • Picture the Problem The period of this physical pendulum is given by

MgDIT π2= where I is the moment of inertia of the circular hoop with respect to an

axis through its pivot point. We can use the parallel-axis theorem to express I in terms of the moment of inertia of the hoop with respect to its center of mass and the distance from its center of mass to its pivot point. Express the period of the physical pendulum: MgD

IT π2=

Using the parallel-axis theorem, find the moment of inertia of the circular hoop about an axis through the pivot point:

2222cm 2MRMRMRMRII =+=+=


gR

MgRMRT 2222

2

ππ ==


( ) s01.2m/s9.81

m0.522 2 == πT

69 • Picture the Problem The period of a physical pendulum is given by

MgDIT π2= where I is its moment of inertia with respect to an axis through its

pivot point. We can solve this equation for I and evaluate it using the given numerical data. Express the period of the physical pendulum: MgD

IT π2=

Solve for I:

2

2

4πMgDTI =

Substitute numerical values and evaluate I:

( )( )( )( )

2

2

22

mkg504.04

s2.6m0.1m/s9.81kg3

⋅=

=π

I

Oscillations

1091

*70 •• Picture the Problem We can use the expression for the period of a simple pendulum to find the period of the clock. (a) Express the period of a simple pendulum: g

T lπ2=


s01.4m/s81.9m42 2 == πT

(b) period. theshortens tray in the

coins placing pendulum, theof mass ofcenter theraisingy effectivelBy

71 •• Picture the Problem Let x be the distance of the pivot from the center of the rod, m the mass at each end of the rod, and L the length of the rod. We can express the period of the physical pendulum as a function of the distance x and then differentiate this expression with respect to x to show that, when x = L/2, the period is a minimum. (a) Express the period of a physical pendulum: MgD

IT π2= (1)

Express the moment of inertia of the dumbbell with respect to an axis through its center of mass:

221

22

cm 22mLLmLmI =⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

Using the parallel-axis theorem, express the moment of inertia of the dumbbell with respect to an axis through the pivot point:

22212

cm 22 mxmLmxII +=+=


xxLC

xxL

g

mgxmxmLT

2241

2241

2221

2

222

+=

+=

+=

π

π

(2)

where g

C π2=

Chapter 14

1092

Set dT/dx = 0 to find the condition for minimum T: extremafor 0

2241

=+

⋅=x

xLdxdC

dxdT

Evaluate the derivative to obtain: ( )

02

2241

2

22412

=+

+−

xxLx

xLx

Because the denominator of this expression cannot be zero, it must be true that:

( ) 02 22412 =+− xLx

Solve for x to obtain: Lx 21=

i.e., the period is a minimum when the pivot point is at one of the masses.

(b) Substitute x = L/4 in equation (2) and simplify to obtain:

( )gL

gLLLT 52

21

2412

41

ππ =+

=


( ) s17.3m/s81.9m25

2 == πT

Remarks: In (a), we’ve shown that x = L/2 corresponds to an extreme value; i.e., to either a maximum or a minimum. To complete the demonstration that this value of x corresponds to a minimum, we can either (1) show that d2T/dx2 evaluated at x = L/2 is positive, or (2) graph T as a function of x and note that the graph is a minimum at x = L/2. 72 •• Picture the Problem Let x be the distance of the pivot from the center of the rod. We’ll express the period of the physical pendulum as a function of the distance x and then differentiate this expression with respect to x to find the location of the pivot point that minimizes the period of the physical pendulum. Express the period of a physical pendulum: MgD

IT π2= (1)

Oscillations

1093

Express the moment of inertia of the dumbbell with respect to an axis through its center of mass:

( )2

32

2121

22

cm 222

mL

LmLmLmI

=

+⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

Using the parallel-axis theorem, express the moment of inertia of the dumbbell with respect to an axis through the pivot point:

2232

2cm

4

4

mxmL

mxII

+=

+=


xxL

g

mgxmxmLT

2232

2232

4

442

+=

+=

π

π

or

xxL

CT22

32 4+

= where g

C π=

Set dT/dx = 0 to find the condition for minimum T: extremafor 04 22

32

=+

×=x

xLdxdC

dxdT

Evaluate the derivative to obtain: ( )0

42

4822

32

2

22322

=+

+−

xxLx

xLx

Because the denominator of this expression cannot be zero, it follows that:

( ) 048 22322 =+− xLx

Solve for x to obtain: 6

Lx =

The distance to the pivot point from the nearer mass is:

LLLd 0918.062=−=

Remarks: We’ve shown that 6Lx = corresponds to an extreme value; i.e., to

either a maximum or a minimum. To complete the demonstration that this value of x corresponds to a minimum, we can either (1) show that d2T/dx2 evaluated at

Chapter 14

1094

6Lx = is positive, or (2) graph T as a function of x and note that the graph is a

minimum at 6Lx = .

*73 •• Picture the Problem Let x be the distance of the pivot from the center of the meter stick, m the mass of the meter stick, and L its length. We’ll express the period of the meter stick as a function of the distance x and then differentiate this expression with respect to x to determine where the hole should be drilled to minimize the period. Express the period of a physical pendulum: MgD

IT π2= (1)

Express the moment of inertia of the meter stick with respect to its center of mass:

2121

cm mLI =

Using the parallel-axis theorem, express the moment of inertia of the meter stick with respect to the pivot point:

22121

2cm

mxmL

mxII

+=

+=


xxLC

xxL

g

mgxmxmLT

22121

22121

22121

2

2

+=

+=

+=

π

π

where g

C π2=

Set dT/dx = 0 to find the condition for minimum T: extremafor 0

22121

=+

×=x

xLdxdC

dxdT


02

22121

2

221212

=+

+−

xxLx

xLx

Because the denominator of this expression cannot be zero, it follows

( ) 02 221212 =+− xLx

Oscillations

1095

that: Solve for and evaluate x to obtain: cm9.28

12cm100

12===

Lx

The hole should be drilled at a distance: cm1.21cm28.9cm50 =−=d

from the center of the meter stick. 74 •• Picture the Problem Let m represent the mass and r the radius of the uniform disk. We’ll use the expression for the period of a physical pendulum and the parallel-axis theorem to obtain a quadratic equation that we can solve for d. We will then treat our expression for the period of the pendulum as an extreme-value problem, setting its derivative equal to zero in order to determine the value for d that will minimize the period. (a) Express the period of a physical pendulum: mgd

IT π2=

Using the parallel-axis theorem, relate the moment of inertia with respect to an axis through the hole to the moment of inertia with respect to the disk’s center of mass:

2221

2cm

mdmR

mdII

+=

+=


gddR

mgdmdmRT

2221

2221

2

2

+=

+=

π

π (1)

Square both sides of this equation, simplify, and substitute numerical values to obtain:

024

2

2

22 =+−

RdgTdπ

or ( ) 0m320.0m553.1 22 =+− dd

Solve the quadratic equation to obtain:

m0.245=d

The second root, d = 1.31 m, is too large to be physically meaningful.

Chapter 14

1096

(b) Set the derivative of equation (1) equal to zero to find relative maxima and minima:

extremafor0

2 2221

=

+⋅=

ddR

ddd

gdddT π


02

222

21

2

22212

=+

+−

ddRd

dRd

Because the denominator of this fraction cannot be zero:

( ) 02 22212 =+− dRd

Solve this equation to obtain: 2

Rd =

Evaluate equation (1) with

2Rd = to obtain an expression

for the shortest possible period of this physical pendulum:

gR

Rg

RRT 22

2

22

212

21

ππ =+

=


( ) s13.2m/s81.9

m8.022 2 == πT

Remarks: We’ve shown that 2Rd = corresponds to an extreme value; i.e., to

either a maximum or a minimum. To complete the demonstration that this value of d corresponds to a minimum, we can either (1) show that d2T/dd2 evaluated at

2Rd = is positive, or (2) graph T as a function of d and note that the graph is a

minimum at 2Rd = .

75 ••• Picture the Problem We can use the equation for the period of a physical pendulum and the parallel-axis theorem to show that h1 + h2 = gT 2/4π 2. Express the period of the physical pendulum: mgd

IT π2=

Using the parallel-axis theorem, relate the moment of inertia with respect to an axis through P1 to the

21cm mhII +=

Oscillations

1097

moment of inertia with respect to the disk’s center of mass: Substitute to obtain:

1

21cm2

mghmhIT +

= π

Square both sides of this equation and rearrange to obtain: 1

1

cm2

2

4mh

hImgT

+=π

(1)

Because the period of oscillation is the same for point P2:

22

cm1

1

cm mhhImh

hI

+=+

Solve this equation for Icm: 21cm hmhI =


11

212

2

4mh

hhmhmgT+=

π

or

2

2

12 4πgThh =+

76 ••• Picture the Problem We can find the period of the physical pendulum in terms of the period of a simple pendulum by starting with mgLIT π2= and applying the parallel-

axis theorem. Performing a binomial expansion for r << L on the radicand of our expression for T will lead to T ≈ T0 (1 + r2/5L2).

(a) Express the period of the physical pendulum:

mgLIT π2=

Using the parallel-axis theorem, relate the moment of inertia of the pendulum about an axis through its center of mass to its moment of inertia with respect to an axis through its point of support:

2252

2cm

mLmr

mLII

+=

+=

Chapter 14

1098


2

2

0

2

2

2

2

225222

52

521

5212

5212

22

LrT

Lr

gL

Lr

gL

gLLr

mgLmLmrT

+=

+=⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

+=

ππ

ππ

(b) Using the binomial expansion,

expand :521

21

2

2

⎟⎟⎠

⎞⎜⎜⎝

⎛+

Lr

2

2

2

2

2

2

221

2

2

51

sorder term-higher 52

81

52

211

521

Lr

Lr

Lr

Lr

+≈

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+=⎟⎟

⎠

⎞⎜⎜⎝

⎛+

provided r << L

Substitute in our result from (a) to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+≈ 2

2

0 51

LrTT

(c) Express the fractional error when the approximation T = T0 is used for this pendulum:

2

2

2

200

0

51

51

1

Lr

Lr

TT

TTT

TT

=−+=

−=−

≈∆

Substitute numerical values and evaluate ∆T/T:

( )( )

0.008%cm1005

cm22

2

=≈∆TT

For an error of 1%:

01.05 2

2

=Lr

Solve for and evaluate r with L = 100 cm:

( )cm22.4

0.05cm10005.0

=

== Lr

77 ••• Picture the Problem The period of this physical pendulum is given by

.2 MgDIT π= We can express its period as a function of the distance d by using the

definition of the center of mass of the pendulum to find D in terms of d and the parallel-axis theorem to express I in terms of d. Solving the resulting quadratic equation yields d.

Oscillations

1099

In (b), because the clock is losing 5 minutes per day, one would reposition the disk so that the clock runs faster; i.e., so the pendulum has a shorter period. We can determine the appropriate correction to make in the position of the disk by relating the fractional time loss to the fractional change in its position.

(a) Express the period of the physical pendulum:

cmtot

2gxmIT π=

Solve forcmxI

: 2

tot2

cm 4πgmT

xI

= (1)

Express the moment of inertia of the physical pendulum, relative to an axis through the pivot point, as a function of d:

22212

312

cm MdMrmLMdII ++=+=


( )( ) ( )( )( )

( ) 22

2

2212

31

kg2.1mkg0802.1kg2.1

m15.0kg2.1m2kg8.0

dd

I

+⋅=

+

+=

Locate the center of mass of the physical pendulum relative to the pivot point:

( ) ( )( ) ( )dx kg1.2m1kg0.8kg2 cm +=

and dx 6.0m4.0cm +=


( ) ( )( ) ( ) 222

2222

m/skg49698.04

kg2m/s81.90.6m4.0

kg2.1mkg0802.1 TTd

d⋅==

++⋅

π (2)

Setting T = 2.5 s and solving for d yields:

m63572.1=d

where we have kept more than three significant figures for use in part (b).

(b) There are 1440 minutes per day. If the clock loses 5 minutes per day, then the period of the clock is related to the perfect period of the clock by:

perfect14401435 TT =

where Tperfect = 3.5 s.

Chapter 14

1100

Solve for and evaluate T:

( )

s51220.3

s5.314351440

14351440

perfect

=

== TT

Substitute T = 3.51220 s in equation (2) and solve for d to obtain:

m40140.3=d

Substitute T = 3.50 s in equation (2) and solve for d ′ to obtain:

m37825.3=d'

Express the distance the disk needs to be moved upward to correct the period:

cm2.32

m37825.3m40140.3

=

−=−=∆ d'dd

*78 •• Picture the Problem The period of a simple pendulum depends on its amplitude φ0

according to ⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛++= ...

21sin

43

21

21sin

2112 0

42

202

2 φφπgLT . We can

approximate T to the second-order term and express ∆T/T = (Tslow – Taccurate)/T. Equating this expression to ∆T/T calculated from the fractional daily loss of time will allow us to solve for and evaluate the amplitude of the pendulum that corresponds to keeping perfect time.

Express the fractional daily loss of time: 86400

48s3600

h1h24

day1day

s48=××=

∆TT

Approximate the period of the clock to the second-order term: ⎥⎦

⎤⎢⎣⎡ += 0

22 2

1sin2112 φπ

gLT

Express the difference in the periods of the slow and accurate clocks:

( )

( )

⎥⎦⎤−

⎢⎣⎡ °=

⎟⎟⎠

⎞⎥⎦⎤

⎢⎣⎡ +−

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ °+=

−=∆

02

2

22

02

2

22

accurateslow

21sin

21

4.821sin

212

21sin

211

4.821sin

2112

φ

π

φ

π

gL

gL

TTT

Oscillations

1101

Divide both sides of this equation by T to obtain: 0

22

21sin

412.4sin

41 φ−°=

∆TT

Substitute for TT∆

and simplify to

obtain:

8640048

21sin

412.4sin

41

022 =−° φ

and

05605.021sin 0 =φ

Solve for φ0: °= 43.60φ

79 •• Picture the Problem The period of a simple pendulum depends on its amplitude φ0

according to ⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛++= ...

21sin

43

21

21sin

2112 0

42

202

2 φφπgLT . We’ll approximate

T to the second-order term and express ∆T/T = (Tslow – Tcorrect)/T. Equating this expression to ∆T/T calculated from the fractional daily loss of time will allow us to solve for and evaluate the amplitude of the pendulum that corresponds to keeping correct time.

Express the fractional daily loss of time: 1440

5min60h1

h24day1

daymin5

=××=∆TT

Approximate the period of the clock to the second-order term: ⎥⎦

⎤⎢⎣⎡ += 0

22 2

1sin2112 φπ

gLT

Assuming that the amplitude of the slow-running clock’s pendulum is small enough to ignore, express the difference in the periods of the slow and corrected clocks:

⎥⎦⎤

⎢⎣⎡−=

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ +−=

−=∆

02

2

02

2

correctslow

21sin

212

21sin

21112

φπ

φπ

gL

gL

TTT

Divide both sides of this expression by T to obtain: 0

2

21sin

41 φ−=

∆TT

Substitute for TT∆

and simplify to

obtain:

14405

21sin

41

02 −

=− φ

and

Chapter 14

1102

1178.021sin 0 =φ

Solve for φ0: °= 5.130φ

Damped Oscillations 80 • Picture the Problem We can use the definition of the damping constant and its dimensions to show that it has units of kg/s.

Using its definition, relate the decay constant τ to the damping constant b: b

m=τ ⇒

τmb =

Substitute the units of m and τ to obtain:

[ ][ ] s

kg lly,Dimensiona ==TMb

81 •

Picture the Problem For small damping, ( )cycle

2EE

Q∆

=π

where ∆E/E is the fractional

energy loss per cycle.

Relate the Q factor to the fractional energy loss per cycle: ( )

cycle

2EE

Q∆

=π

Solve for and evaluate the fractional energy loss per cycle:

( ) %14.320022

cycle===∆

ππQ

EE

82 • Picture the Problem We can find the period of the oscillator from kmT π2= and its

total initial energy from 221

0 kAE = . The Q factor can be found from its definition

( )cycle

2 EEQ ∆= π and the damping constant from .0 bmQ ω=

(a) The period of the oscillator is given by: k

mT π2=


N/m400kg22 == πT

(b) Relate the initial energy of the 2

21

0 kAE =

Oscillations

1103

oscillator to its amplitude:

Substitute numerical values and evaluate E0:

( )( ) J0.180m0.03N/m400 221

0 ==E

(c) Relate the fractional rate at which the energy decreases to the Q value and evaluate Q:

( ) 62801.0

22

cycle

==∆

=ππ

EEQ

Express the Q value in terms of b:

bmQ 0ω=

Solve for the damping constant b:

TQm

Qmb πω 20 ==

Substitute numerical values and evaluate b:

( )( )( ) kg/s0451.0

628s444.0kg22

==πb

83 •• Picture the Problem The amplitude of the oscillation at time t is ( ) τ2

0teAtA −= where

τ = m/b is the decay constant. We’ll express the amplitudes one period apart and then show that their ratio is constant.

Relate the amplitude of a given oscillation peak to the time at which the peak occurs:

( ) τ20

teAtA −=

Express the amplitude of the oscillation peak at t′ = t + T:

( ) ( ) τ20

TteATtA +−=+

Express the ratio of these consecutive peaks:

( )( ) ( )

constant

22

0

20

=

==+

−+−

−τ

τ

τT

Tt

t

eeA

eATtA

tA

84 •• Picture the Problem We can relate the fractional change in the energy of the oscillator each cycle to the fractional change in its amplitude. Both the Q value and the decay constant τ can be found from their definitions.

(a) Relate the energy of the oscillator to its amplitude:

221 kAE =

Chapter 14

1104

Take the differential of this relationship to obtain:

kAdAdE =

Divide both sides of this equation by E: A

dAkA

kAdAE

dE 2221

==

Approximate dE and dA by ∆E and ∆A and evaluate ∆E/E:

%10%)5(2 ==∆EE

(b) For small damping:

τT

EE

=∆

and

s300.01

s3==

∆=

EETτ

(c) Using its definition, express and evaluate Q:

( ) 8.62s30s3

220 ====

πτπτωT

A

85 •• Picture the Problem We can use the physical interpretation of Q for small damping

( )cycle

2EE

Q∆

=π

to find the fractional decrease in the energy of the oscillator each

cycle.

(a) Express the fractional decrease in energy each cycle as a function of the Q factor and evaluate EE∆ :

314.02022

===∆ ππ

QEE

(b) Using the definition of the Q factor, use Equation 14-35 to express ω′ as a function of Q:

21

20

21

20

2

2

0

411

411

⎥⎦

⎤⎢⎣

⎡−=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

Q

mb'

ω

ωωω

Use the approximation (1 + x)½ ≈ 1 + ½x for small x to obtain: ⎥

⎦

⎤⎢⎣

⎡−= 20 8

11Q

' ωω

Oscillations

1105

Express and evaluate ω′ − ω0:

( )percent1013.3

2081

81

811

2

2

20200

−×−=

−=

−=−⎥⎦

⎤⎢⎣

⎡−=−

QQ' ωωωω

86 •• Picture the Problem The amplitude of the spring-and-mass oscillator varies with time according to τ2

0teAA −= and its energy according to τteEE −= 0 .

(a) Express the amplitude of the oscillations as a function of time:

( ) s4cm6 teA −=

Evaluate the amplitude when t = 2 s: ( ) ( ) ( )cm64.3

cm6cm6s2 21s4s2

=

== −− eeA

Evaluate the amplitude when t = 4 s: ( ) ( ) ( )

cm21.2

cm6cm6s4 1s4s4

=

== −− eeA

(b) Express the energy of the system at t = 0:

( ) J600 0s20

0 === − EeEE

Express the energy in the system at t =2 s:

( ) 10

s2s20s2 −− == eEeEE

The energy dissipated in the first 2 s is: ( ) ( )( )

( )( )J9.37

1J60

1

s20

1

10

s20

=

−=

−=

−=∆

−

−

−

e

eE

EEE

The energy dissipated in the second 2-s interval is:

( )( )( ) J0.241J9.37

11

s22ss2s42

=−=

−=∆−

−−

e

eEE

*87 •• Picture the Problem We can find the fractional loss of energy per cycle from the physical interpretation of Q for small damping. We will also find a general expression for the earth’s vibrational energy as a function of the number of cycles it has completed. We can then solve this equation for the earth’s vibrational energy after any number of days.

Chapter 14

1106

(a) Express the fractional change in energy as a function of Q:

%57.140022

===∆ ππ

QEE

(b) Express the energy of the damped oscillator after one cycle:

⎟⎠⎞

⎜⎝⎛ ∆−=

EEEE 101

Express the energy after two cycles: 2

012 11 ⎟⎠⎞

⎜⎝⎛ ∆−=⎟

⎠⎞

⎜⎝⎛ ∆−=

EEE

EEEE

Generalizing to n cycles: ( )

( )n

nn

n

E

EEEEE

9843.0

0157.011

0

00

=

−=⎟⎠⎞

⎜⎝⎛ ∆−=

(c) Express 2 d in terms of the number of cycles; i.e., the number of vibrations the earth will have experienced:

T

T

3.53min54

1min2880

hm60

dh24d2d2

=

×=

××=

Evaluate E(2 d): ( ) 0

3.530 430.0)9843.0(d2 EEE ==

88 •• Picture the Problem The diagram shows 1) the pendulum bob displaced through an angle θ0 and held in equilibrium by the force exerted on it by the air from the fan and 2) the bob accelerating, under the influence of gravity, tension force, and drag force, toward its equilibrium position. We can apply Newton’s 2nd law to the bob to obtain the differential equation of motion of the damped pendulum and then use its solution to find the decay time constant and the time required for the amplitude of oscillation to decay to 1°. (a) Apply ∑ = ατ I to the pendulum to obtain:

2

2

dsindtdIFmg θθ =+− ll

Oscillations

1107

Express the moment of inertia of the pendulum with respect to an axis through its point of support:

2lmI =

Substitute for I and Fd to obtain: 0sin2

22 =++ θθ

lll mgbvdtdm

Because θ << 1 and v = lω = ldθ/dt: 02

2

22 =++ θθθ

lll mgdtdb

dtdm

or

02

2

=++ θθθl

mgdtdb

dtdm

The solution to this second-order homogeneous differential equation with constant coefficients is:

( )δωθθ τ += − 'te t cos20 (1)

where θ0 is the maximum amplitude, τ = m/b is the time constant, and the

frequency ( )200 21 ωωω mb' −= .

Apply aF rrm=∑ to the bob when

it is at its maximum angular displacement to obtain:

0sin 0fan =−=∑ θTFFx and

0cos 0 =−=∑ mgTFy θ

Divide the x equation by the y equation to obtain:

00

0fan tancossin θ

θθ

==TT

mgF

or 0fan tanθmgF =

When the bob is in equilibrium, the drag force on it equals Ffan:

0tanθmgbv =

Solve for m/b in the definition of τ to obtain:

0tanθτ

gv

bm==

Substitute numerical values and evaluate τ : ( ) s16.8

5tanm/s9.81m/s7

2 =°

=τ

(b) From equation (1) we have: τθθ 2

0te−=

When the amplitude has decreased to 1°:

°=° − 15 2τte or 2.02 =− τte

Take the natural logarithm of both sides of the equation to obtain:

( )2.0ln2

=−τt

Chapter 14

1108

Solve for t: ( )2.0ln2τ−=t

Substitute for τ and evaluate t: ( ) ( ) s3.262.0lns16.82 =−=t

Driven Oscillations and Resonance 89 • Picture the Problem The resonant frequency of a vibrating system depends on the mass

of the system and on a “stiffness” constant according to mkf

π21

0 = or, in the case of a

simple pendulum oscillating with small-amplitude vibrations, .21

0 Lgf

π=

(a) For this spring-and-mass oscillator we have: Hz01.1

kg10N/m400

21

0 ==π

f

(b) For this spring-and-mass oscillator we have: Hz01.2

kg5N/m800

21

0 ==π

f

(c) For this simple pendulum we have:

Hz352.0m2m/s9.81

21 2

0 ==π

f

90 • Picture the Problem We can use the physical interpretation of Q for small damping to find the Q factor for this damped oscillator. The width of the resonance curve depends on the Q factor according to .0 Qωω =∆

(a) Using the physical interpretation of Q for small damping, relate Q to the fractional loss of energy of the damped oscillator per cycle:

( )cycle

2EE

Q∆

=π

Evaluate this expression for ( ) %2

cycle=∆ EE :

31402.0

2==

πQ

(b) Relate the width of the resonance curve to the Q value of the oscillatory system:

Qf

Q00 2πωω ==∆

Oscillations

1109

Substitute numerical values and evaluate ∆ω:

( ) rad/s00.63.14

s3002 -1

==∆πω

91 •• Picture the Problem The amplitude of the damped oscillations is related to the damping constant, mass of the system, the amplitude of the driving force, and the natural and

driving frequencies through( ) 22222

02

0

ωωω bm

FA+−

= . Resonance occurs when

.0ωω = At resonance, the amplitude of the oscillations is 220 ωbFA = and the

width of the resonance curve is related to the damping constant and the mass of the system according to .mb=∆ω

(a) Express the amplitude of the oscillations as a function of the driving frequency:

( ) 222220

2

0

ωωω bm

FA+−

=

Determine ω0: rad/s14.14kg2N/m400

0 ===mkω

Evaluate the radicand in the expression for A to obtain:

( ) ( ) ( )[ ]( ) ( )

424

22

2222

s/kg1004.4rad/s10kg/s2

rad/s10rad/s14.14kg2

×=

+

−


cm98.4s/kg1004.4

N10424=

×=A

(b) Resonance occurs when: rad/s1.140 == ωω

(c) Express the amplitude of the motion at resonance: 2

020

ωbFA =

Substitute numerical values and evaluate A: ( ) ( )

cm35.4rad/s14.14kg/s2

N1022==A

(d) The width of the resonance curve is:

rad/s00.1kg2

kg/s2===∆

mbω

Chapter 14

1110

92 •• Picture the Problem We’ll find a general expression for the damped oscillator’s energy as a function of the number of cycles it has completed. We can then solve this equation for the number of cycles corresponding to the loss of half the oscillator’s energy. The Q factor is related to the fractional energy loss per cycle through QEE π2=∆ and the width of the resonance curve is Q0ωω =∆ where ω0 is the oscillator’s natural angular

frequency.

(a) Express the energy of the damped oscillator after one cycle:

⎟⎠⎞

⎜⎝⎛ ∆−=

EEEE 101

Express the energy after two cycles: 2

012 11 ⎟⎠⎞

⎜⎝⎛ ∆−=⎟

⎠⎞

⎜⎝⎛ ∆−=

EEE

EEEE

Generalizing to n cycles: n

n EEEE ⎟⎠⎞

⎜⎝⎛ ∆−= 10

Substitute numerical values: ( )nEE 035.015.0 00 −=

or ( )n965.05.0 =

Solve for n to obtain:

cycles. complete 20

5.19965.0ln

5.0ln

≈

==n

(b) Apply the physical interpretation of Q for small damping to obtain:

180035.022

==∆

=ππ

EEQ

(c) The width of the resonance curve is given by:

( )

rad/s49.3

180Hz10022 00

=

===∆ππωω

Qf

Q

Collisions 93 ••• Picture the Problem Let the system include the spring-and-mass oscillator and the second object of mass m. Because the net external force acting on this system is zero, momentum is conserved during the collision of the second object with the oscillator.

Oscillations

1111

Because the collision is elastic, we can also apply conservation of energy. Let the subscript 1 refer to the object attached to the spring and the subscript 2 identify the second object. (a) Using momentum conservation, relate the speeds of the objects before and after their collision:

2f2i1i mvmvmv =+

or 2f2i1i vvv =+ (1)

Using conservation of energy, obtain a second relationship between the speeds of the objects before and after their collision:

2f22

12i22

12i12

1 mvmvmv =+

or 2f2

2i2

2i1 vvv =+ (2)

Solve equation (2) for :2

i2v ( )( )1i2f1i2f2i1

2f2

2i2 vvvvvvv −+=−=

Substitute for v2f from equation (1): ( )( )

( )( ) 2i22i1i2i2i1i

1i2i1i1i2i1i2i2

22 vvvvvv

vvvvvvv

+=+=

−+++=

or 02 2i1i =vv

Because 01i ≠v , it follows that: 02i == vv

i.e., the second object must be initially at rest.

(b) Because v2i = 0, we have, from equation (1):

1i2f vv =

Because the object connected to the spring was moving through its equilibrium position at the time of collision:

( )( )m/s00.4

s40m1.0 1max1i

=

=== −ωAvv

94 ••• Picture the Problem Let the system include the spring-and-mass oscillator and the second object of mass m. Because the net external force acting on this system is zero, momentum is conserved during the collision of the second object with the oscillator. Because the collision is elastic, we can also apply conservation of energy. Let the subscript 1 refer to the object attached to the spring and the subscript 2 identify the second object.

Using momentum conservation, relate the speeds of the objects

2f2i1i mvmvmv =+

or

Chapter 14

1112

before and after their collision:

2f2i1i vvv =+ (1)

Using conservation of energy, obtain a second relationship between the speeds of the objects before and after their collision:

2f22

12i22

12i12

1 mvmvmv =+

or 2f2

2i2

2i1 vvv =+ (2)

Solve equation (2) for :2

i2v ( )( )1i2f1i2f2i1

2f2

2i2 vvvvvvv −+=−=

Substitute for v2f from equation (1): ( )( )

( )( ) 2i22i1i2i2i1i

1i2i1i1i2i1i2i2

22 vvvvvv

vvvvvvv

+=+=

−+++=

or 02 2i1i =vv

Because 01i ≠v , it follows that: 02i == vv

i.e., the second object must be initially at rest.

Because the object connected to the spring was moving through its equilibrium position at the time of collision:

( )( )m/s4

s40m1.0 1max1i

==== −ωAvv

Express the total energy of the system just before the collision:

2i12

1 mvE =

Solve for m: 2i1

2vEm =


( )( )

kg00.1m/s4

J822 ==m

Relate the spring constant to the angular frequency of the oscillator:

2ωmk =


( )( ) kN/m60.1s40kg1 21 == −k

95 ••• Picture the Problem Let the system include the spring-and-mass oscillator and the 1-kg object. Because the net external force acting on this system is zero, momentum is

Oscillations

1113

conserved during the collision of the second object with the oscillator. Let the subscript 1 refer to the 1-kg object and the subscript 2 to the 2-kg object. We can relate the amplitude of the motion to the maximum speed of the oscillator (which we can find from conservation of momentum) and the angular frequency of the oscillator, which we can determine from its definition. Once we have found the amplitudes and angular frequencies for both collisions, we express the position of each as a function of time, using the initial conditions to find the phase constants.

(a) Relate the amplitude of the motion to the angular frequency and maximum speed of the oscillator:

ωmaxvA = (1)

Because the 2-kg object is initially at rest, the maximum speed of the oscillator will be its speed immediately after the collision. Use conservation of momentum to relate this maximum speed to the speed of the 1-kg object before the collision:

( ) max211i1 vmmvm +=

Solve for vmax: 1i

21

1max v

mmmv+

=


( ) m/s2m/s6kg2kg1

kg1max =

+=v

Express the angular frequency of the oscillator:

21 mmk+

=ω


kg3N/m600

==ω

Substitute in equation (1) and evaluate A:

cm1.14s14.14

m/s21 == −A

Express and evaluate the period of the oscillator’s period:

s444.0s14.14

221 === −

πωπT

(b) For an elastic collision: 1i

21

12fmax

2 vmm

mvv+

==

Chapter 14

1114


( ) ( ) m/s4m/s6kg3kg12

max ==v

Using its definition, evaluate the angular frequency of the oscillator:

rad/s17.32kg2N/m600

2

===mkω

Substitute in equation (1) and evaluate A:

cm1.23s17.32

m/s41 == −A

Express and evaluate the period of the oscillator’s period:

s363.0s32.17

221 === −

πωπT

(c) For the perfectly inelastic collision:

( ) ( ) ( )[ ]δ+= − ttx 1s1.14coscm1.14 (2)

Use the initial conditions to evaluate δ: ( ) 20

tantan 01

0

01 πωω

δ −=⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−= −− v

xv

Substitute in equation (2) to obtain: ( ) ( ) ( )

( ) ( )[ ]tttx

1

1

s1.14sincm1.14

2s1.14coscm1.14

−

−

=

⎥⎦⎤

⎢⎣⎡ −=

π

For the elastic collision:

( ) ( ) ( )[ ]δ+= − ttx 1s3.17coscm1.23 (3)

Use the initial conditions to evaluate δ: ( ) 20

tantan 01

0

01 πωω

δ −=⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−= −− v

xv


( ) ( )[ ]tttx

1

1

s3.17sincm1.23

2s3.17coscm1.23

−

−

=

⎥⎦⎤

⎢⎣⎡ −=

π

General Problems 96 • Picture the Problem The particle’s displacement is of the form ( )δω += tAx cos . Thus,

we have A = 0.4 m, ω = 3 rad/s, and δ = π/4. We can find the frequency of the motion from its angular frequency and the period from the frequency. The particle’s position at t = 0 and t = 0.5 s can be found directly from its displacement function.

Oscillations

1115

(a) Express and evaluate the frequency of the particle’s motion:

Hz477.02rad/s3

2===

ππωf

Use the relationship between the frequency and the period of the particle’s motion to find its period:

s2.09s0.477

111 === −f

T

(b) Using the expression for the particle’s displacement, find its position at t = 0:

( ) ( ) ( )( )

( ) m283.04

cosm4.0

40rad/s3cosm4.00

=⎥⎦⎤

⎢⎣⎡=

⎥⎦⎤

⎢⎣⎡ +=

π

πx

(c) Using the expression for the particle’s displacement, find its position at t = 0.5 s:

( ) ( ) ( )( )

( ) [ ]m264.0

rad29.2cosm4.04

s5.0rad/s3cosm4.00

−=

=

⎥⎦⎤

⎢⎣⎡ +=

πx

97 • Picture the Problem We can express the velocity of the particle by differentiating its displacement with respect to time. (a) Differentiate the particle’s displacement to obtain:

( ) ( )

( ) ( ) ⎥⎦⎤

⎢⎣⎡ +−=

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ +=

=

4rad/s3sinm/s2.1

4rad/s3sinm4.0

π

π

t

tdtddtdxv

(b) Evaluate the result in part (a) at t = 0:

( ) ( ) ( )( )

( )

m/s849.0

4sinm/s2.1

40rad/s3sinm/s2.10

−=

⎥⎦⎤

⎢⎣⎡−=

⎥⎦⎤

⎢⎣⎡ +−=

π

πv

(c) By inspection of the result in part (a) (or from ωAv =max ):

m/s20.1max =v

(d) Substitute vmax for v to obtain: ( ) ( ) ⎥⎦

⎤⎢⎣⎡ +−=

4'rad/s3sinm/s2.1m/s2.1 πt

Chapter 14

1116

or

( ) ( )2

31sin4

rad/s3 1 ππ=−=+ −t'

Solve for t′ to obtain: s31.1=t'

98 • Picture the Problem Let ∆y represent the amount by which the spring stretches. We’ll apply a condition for equilibrium to the object to relate the amount the spring has stretched to the angular frequency of its motion and then solve this equation for ∆y. Apply 0=∑i yF to the object

when it is in its equilibrium position and solve for the elongation of the spring:

0=−∆ mgyk

or

2ωgg

kmy ==∆

Relate the angular frequency of the object’s motion to its period: T

πω 2=


gTy2

2⎟⎠⎞

⎜⎝⎛=∆π

Substitute numerical values and evaluate ∆x: ( ) m03.5m/s81.9

2s5.4 2

2

=⎟⎠⎞

⎜⎝⎛=∆

πy

*99 •• Picture the Problem Compare the forces acting on the particle to the right in Figure 14-36 with the forces shown acting on the bob of the simple pendulum shown in the free-body diagram to the right. Because there is no friction, the only forces acting on the particle are mg and the normal force acting radially inward. In (b), we can think of the particles as the bobs of simple pendulums of equal length.

(a) The normal force is identical to the tension in a string of length r that keeps the particle moving in a circular path and a component of mg provides, for small displacements θ0 or s2, the linear restoring force required for oscillatory motion.

Oscillations

1117

(b) The particles meet at the bottom. Because s1 and s2 are both much smaller than r, the particles behave like the bobs of simple pendulums of equal length; therefore they have the same periods. 100 •• Picture the Problem The diagram shows the ball when it is a horizontal distance x from the bottom of the bowl. Note that we’ve chosen the zero of gravitational potential energy to be at the bottom of the bowl. The total energy of the ball is the sum of its potential energy and kinetic energies due to translation and rotation. Once we’ve obtained an expression for the total energy of the rolling ball, we can require, because the surface is frictionless, that the total energy of the sliding object be the same as that of the rolling ball. Because the motion of the ball is simple harmonic motion, we can assume a solution to its differential equation of motion and express the total energy of the ball in terms of this assumed solution. Doing so will lead us to an expression that we can solve for the oscillation frequency of the ball.

(a) Express the total energy E of the ball:

rottrans KKUKUE ++=+= (1)

Referring to the diagram shown above and assuming that R << r, express the potential energy of the ball when it is a horizontal distance x from the bottom of the bowl:

( ) ( )θcos1−= mgrxU

Express cosθ as a power series: ...

!4!21cos

42

++−=θθθ

For θ << 1:

!21cos

2θθ −≈


( ) 221

2

!211 θθ mgrmgrxU =⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−≈

Chapter 14

1118

For R << r:

rx

≈θ


( )r

mgxxU2

2

=


222

21

21

2ωImv

rmgxE ++=

Because the ball is rolling without slipping, v = Rω. Substitute for ω and I to obtain:

222

2

52

21

21

2⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛++=

RvmRmv

rmgxE


2

107

2mv

rmgxE +=

(b) Because energy is conserved if the side of the bowl is frictionless:

constant107

22

2

=+= mvr

mgxE

Because the motion is simple harmonic motion, assume a solution of the form:

( )δω += txx cos.0

Differentiate this assumed solution with respect to time to obtain:

( )δωω +−= txv sin0

Substitute to obtain: ( )( )

( )( )

( )

( )δωω

δω

δωω

δω

++

+=

+−+

+=

txm

tr

mgx

txm

txr

mgE

220

2

220

20

2.0

sin10

7

cos2

sin107

cos2

Express the condition the E = constant: 10

72

20

220 xm

rmgx ω

= or 5

7 2ω=

rg


rg

75

=ω

Oscillations

1119

101 •• Picture the Problem Assume that the plane is accelerating to the right with an acceleration a0. The free-body diagram shows the forces on the bob as seen in the accelerated frame of the airplane. Let g′ represent the effective value of the acceleration due to gravity. The period of the yo-yo is given by

g'LT π2=

where g′ is the effective value of the acceleration due to gravity.

Express the period of your yo-yo pendulum as a function of the effective value for the acceleration due to gravity:

g'LT π2=

Using the FBD, relate g′ and g: θcosmg'mg = ⇒ θcos

gg' =


gLT θπ cos2=


( ) s62.1m/s81.9

22cosm7.02 2 =°

= πT

102 •• Picture the Problem The diagram shows the wire described in the problem statement with an object of moment of inertia I suspended from its end. We can apply Newton’s 2nd law to the suspended object to obtain its differential equation of motion. By comparing this equation to the equation of a simple harmonic oscillator, we can show that .Iκω = Apply ατ I=∑ to the object hung from the wire to obtain: 2

2

dtdII θακθ ==−

Chapter 14

1120

Divide both sides of this differential equation by I to obtain: 02

2

=+ θκθIdt

d

This equation can be written as:

022

2

=+ θωθdtd

where Iκω =

103 •• Picture the Problem The diagram shows the torsion balance described in the problem statement. We can apply Newton’s 2nd law to the suspended object to obtain its differential equation of motion. By comparing this equation and its solution to that of a simple harmonic oscillator, we can obtain an equation that we can solve for the torsion constant κ.

Apply ατ I=∑ to the torsion pendulum:

2

2

dtdII θακθ ==−

or

02

2

=+ θκθIdt

d (1)

The differential equation of simple harmonic motion is: 02

2

2

=+ xdt

xd ω

where

( ) ( )δω += txtx cos0 and Tπω 2

=

The solution to equation (1) is:

( )δωθθ += tt cos)( 0 where

Iκω =

Solve for κ to obtain:

I2ωκ =

Oscillations

1121

Express the moment of inertia of the torsion pendulum: 22

222ll mmI =⎟

⎠⎞

⎜⎝⎛=


2

22

2

2222 22

42 T

mTmm lll ππωκ ===

Substitute numerical values and evaluate κ:

( )( )( )

m/radN1086.3

s80m05.0kg050.02

7

2

22

⋅×=

=

−

πκ

*104 •• Picture the Problem Choose a coordinate system in which the direction the cube is initially displaced (downward) is the positive y direction. The figure shows the forces acting on the cube when it is in equilibrium floating in the water and when it has been pushed down a small distance y. We can find the period of its oscillatory motion from its angular frequency. By applying Newton’s 2nd law to the cube, we can obtain its equation of motion; from this equation we can determine the angular frequency of the cube’s small-amplitude oscillations.

Express the period of oscillation in terms of the angular frequency of the oscillations:

ωπ2

=T (1)

Apply ∑ = 0yF to the cube when

it is floating in the water:

0B =− Fmg

Apply ∑ = yy maT to the cube

when it is pushed down a small distance y:

y' maFmg =− B

Chapter 14

1122

Eliminate mg between these equations to obtain:

y' maFF =− BB

or

y' maFFF =−=∆ BBB

For y << 1:

2

22

BB dtydmgyaVgdFF =−=−=≈∆ ρρ

Rewrite the equation of motion as:

gyadt

ydm ρ22

2

−=

or

yym

gadt

yd 22

2

2

ωρ−=−=

where m

ga ρω2

2 =

Solve for ω:

mga ρω =


gm

amga

Tρ

πρπ 22

==

105 •• Picture the Problem Assume that the density of the earth ρ is constant and let m represent the mass of the clock. We can decide the question of where the clock is more accurate by applying the law of gravitation to the clock at a depth h below/above the surface of the earth and at the earth’s surface and expressing the ratios of the acceleration due to gravity below/above the surface of the earth to its value at the surface of the earth. Express the gravitational force acting on the clock when it is at a depth h in a mine:

( )2E hRGM'mmg'−

=

where M′ is the mass between the location of the clock and the center of the earth.

Express the gravitational force acting on the clock at the surface of the earth:

2E

E

RmGMmg =

Oscillations

1123

Divide the first of these equations by the second to obtain: ( )

( )2E

2E

E2E

E

2E

hRR

MM'

RGM

hRGM'

gg'

−=−=

Express M ′:

( )3E34 hRV'M' −== πρρ

Express ME: 3

E34

E RVM πρρ ==


( )2E

2E

3E3

4

3E3

4

hRR

RhR

gg'

−−

=πρ

πρ

Simplify and solve for g′:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=

EE

E 1Rhg

RhRgg'

or

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

E

1Rhgg' (1)

Express the gravitational force acting on the clock when it is at an elevation h:

( )2E

E

hRmGMmg''

+=

Express the gravitational force acting on the clock at the surface of the earth:

2E

E

RmGMmg =


( )

2

E

2E

2E

2E

E

2E

E

1

1

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

−=+=

Rh

hRR

RGM

hRGM

gg''

Solve for g′′: 2

E

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

Rhgg''

Chapter 14

1124

elevated. isclock theifgreater iserror the Thus, . is than closer to is that see we(2), and (1) equations Comparing g''gg'

106 •• Picture the Problem The figure shows this system when it has an angular displacement θ. The period of the system is related to its angular frequency according to T = 2π/ω. We can find the equation of motion of the system by applying Newton’s 2nd law. By writing this equation in terms of θ and using a small-angle approximation, we’ll find an expression for ω that we can use to express T.

(a) Express the period of the system in terms of its angular frequency:

ωπ2

=T (1)

Apply ∑ = aF rrm to the bob: ∑ =−−= xx MaTkxF θsin

and

∑ =−= 0cos MgTFy θ

Eliminate T between the two equations to obtain:

xMaMgkx =−− θtan

Noting that x = Lθ and

,2

2

dtdLLaxθα ==

eliminate the variable x in favor of θ :

θθθ tan2

2

MgkLdtdML −−=

For θ << 1, tanθ ≈ θ :

( )θ

θθθ

MgkL

MgkLdtdML

+−=

−−=2

2

or

θωθθ 22

2

−=⎟⎠⎞

⎜⎝⎛ +−=

Lg

Mk

dtd

where

Lg

Mk+=ω

Oscillations

1125


Lg

Mk

T+

=π2

(b) When T = 2 s and M = 1 kg we have: L

gπ22 =

When T = 1 s we have:

Lgk +

=π21


N/m6.29=k

107 •• Picture the Problem Applying Newton’s 2nd law to the first object as it is about to slip will allow us to express µs in terms of the maximum acceleration of the system which, in turn, depends on the amplitude and angular frequency of the oscillatory motion.

(a) Apply ∑ = xx maF to the

second object as it is about to slip:

max2maxs, amf =

Apply ∑ = 0yF to the second

object:

02n =− gmF

Use nsmaxs, Ff µ= to eliminate

maxs,f and nF between the two

equations:

max22s amgm =µ

and

gamax

s =µ

Relate the maximum acceleration of the oscillator to its amplitude and angular frequency:

21

2max mm

kAAa+

== ω

Substitute for amax to obtain: ( )gmm

Ak

21s +=µ

Chapter 14

1126

(b)increased. is and system theof mass total theincreasing

by reduced is . because unchanged is unchanged. is 221

TkAEEA ω=

108 •• Picture the Problem The diagram shows the box hanging from the stretched spring and the free-body diagram when the box is in equilibrium. We can apply ∑ = 0yF to the box to derive an expression for x. In (b) and (c), we can proceed similarly to obtain expressions for the effective spring constant, the new equilibrium position of the box, and frequency of oscillations when the box is released.

(a) Apply ∑ = 0yF to the box to obtain:

( ) 00 =−− mgxxk

Solve for x: 0x

kmgx +=


( )( )

m46.2

m5.0N/m500

m/s81.9kg100 2

=

+=x

(b) Draw the free-body diagram for the block with the two springs exerting equal upward forces on it:

Apply ∑ = 0yF to the box to obtain:

( ) ( ) 000 =−−+− mgxxkxxk or

( ) 00eff =−− mgxxk (1) where

kk 2eff =

Oscillations

1127

When the box is displaced from this equilibrium position and released, its motion is simple harmonic motion and its frequency is given by:

mk

mk 2eff ==ω


( ) rad/s16.3kg100N/m5002

==ω

(c) Solve equation (1) for x:

02x

kmgx +=


( )( )( )m48.1

m5.0N/m5002

m/s81.9kg100 2

=

+=x

109 •• Picture the Problem We’ll differentiate the expression for the period of simple

pendulum gLT π2= with respect to g, separate the variables, and use a differential

approximation to establish that .21

gg

TT ∆

−≈∆

(a) Express the period of a simple pendulum in terms of its length and the local value of the acceleration due to gravity:

gLT π2=

Differentiate this expression with respect to g to obtain:

[ ]

gT

gLgLdgd

dgdT

2

2 2321

−=

−== −− ππ

Separate the variables to obtain: g

dgTdT

21

−=

Approximate dT and dg by ∆T and ∆g for ∆g << g:

gg

TT ∆

−≈∆

21

(b) Solve the result in part (a) for ∆g: T

Tgg ∆−=∆ 2

Chapter 14

1128

Express ∆T/T:

31004.1s3600

h1h24

d1ds90

−×−=

××−=∆TT

Substitute and evaluate ∆g: ( )( )

22

32

cm/s2.04m/s.02040

1004.1m/s81.92

==

×−−=∆ −g

110 •• Picture the Problem We can find the frequency of the vibrating system from its angular frequency; this depends on the spring constant and the total mass involved in the motion. The energy of the system can be found from the amplitude of its motion. (a) Relate the frequency of the vibrating system to its angular frequency:

mkf

221

2 ππω

==

Substitute numerical values and evaluate f: ( ) Hz25.2

kg6.02N/m240

21

==π

f

Express the total energy of the system:

221 kAE =


( )( ) J43.2m0.6N/m240 221 ==E

(b) (1) The glue dissolves when the spring is at maximum compression:

Relate the frequency to the system’s new angular frequency:

mkf

ππω

21

21

1 ==

Substitute numerical values and evaluate f1:

Hz18.3kg6.0N/m240

21

1 ==π

f

Express the system’s new amplitude as a function of the oscillator’s maximum speed and its new angular frequency:

kmvvA max

1

max1 ==

ω

Find the maximum speed of the oscillator:

( )( )m/s48.8

m0.6s25.222 1max

==== −ππω fAAv

Oscillations

1129

Substitute and evaluate A1: ( )

cm42.4

N/m240kg0.6m/s8.481

=

=A

Express and evaluate the energy of the system:

( )( )J21.6

m0.424N/m240 2212

121

1

=

== kAE

(b) (2) The glue dissolves when the spring is at maximum extension and f2 is the same as f1:

Hz18.32 =f

Because the second object is at rest, the amplitude and energy of the system are unchanged:

m600.02 == AA

and J43.22 == EE

111 •• Picture the Problem Choose a coordinate system in which the positive x direction is to the right and assume that the object is displaced to the right. In case (a), note that the two springs undergo the same displacement whereas in (b) they experience the same force. (a) Express the net force acting on the object:

( ) xkxkkxkxkF eff2121net −=+−=−−=

where 21eff kkk +=

(b) Express the force acting on each spring and solve for x2:

2211 xkxkF −=−=

or

12

12 x

kkx =

Express the total extension of the springs: eff

21 kFxx −=+

Solve for keff:

211

2

11

11

21

11

21eff

111

kkx

kkx

xkxxxk

xxFk

+=

+=

+−

−=+

−=

Chapter 14

1130

Take the reciprocal of both sides of the equation to obtain:

21eff

111kkk

+=

*112 •• Picture the Problem If the displacement of the block is y = A sin ωt, its acceleration is a = −ω2Asinωt. (a) At maximum upward extension, the block is momentarily at rest. Its downward acceleration is g. The downward acceleration of the piston is ω 2A. Therefore, if ω 2A > g, the block will separate from the piston. (b) Express the acceleration of the small block:

tAa ωω sin2−=

For gA 32 =ω and A = 15 cm: gtga −=−= ωsin3

Solve for t:

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛= −−

31sin

331sin1 11

gAt

ω

Substitute numerical values and evaluate t: ( ) s0243.0

31sin

m/s81.93m15.0 1

2 == −t

113 •• Picture the Problem The plunger and ball are moving with their maximum speed as they pass through their equilibrium position (x = 0). Once it has passed its equilibrium position, the acceleration of the plunger becomes negative; therefore it begins to slow down and the ball, continuing with speed vs, separates from the plunger. We can find this separation speed by equating it to the maximum speed of the plunger. Application of conservation of energy to the motion of the plunger will allow us to express the distance at which the plunger comes momentarily to rest. (a) The ball will leave the plunger when the plunger is moving with its maximum speed; i.e., at its equilibrium position:

0=x

(b) Express the speed of the ball upon separation in terms of the maximum speed of the plunger:

ωω 0maxs xAvv ===

Oscillations

1131

The angular frequency is given by:

pb mmk+

=ω


pb0s mm

kxv+

=

(c) Apply conservation of energy to the plunger:

0si,sf,if =−+− UUKK

or, because Kf = Ui = 0, 02

f212

sp21 =+− kxvm

Solve for xf:

sp

f vk

mx =

Substitute for vs and simplify to obtain:

pb

p0f mm

mxx

+=

114 •• Picture the Problem Applying Newton’s 2nd law to the box as it is about to slip will allow us to express µs in terms of the maximum acceleration of the platform which, in turn, depends on the amplitude and angular frequency of the oscillatory motion. (a) Apply ∑ = xx maF to the box

as it is about to slip:

maxmaxs, maf =

Apply ∑ = 0yF to the box: 0n =− mgF

Use nsmaxs, Ff µ= to eliminate

maxs,f and nF between the two

equations:

maxs mamg =µ

and

gamax

s =µ

Relate the maximum acceleration of the oscillator to its amplitude and angular frequency:

2max ωAa =

Substitute for amax : gTA

gA

2

22

s4πωµ ==

Chapter 14

1132

Substitute numerical values and evaluate µs:

( )( ) ( ) 52.2

m/s9.81s8.0m0.44

22

2

s ==πµ

(b) Solve the equation derived above for Amax: 2

2s

2s

max 4πµ

ωµ gTgA ==

Substitute numerical values and evaluate Amax:

( )( )( )

cm36.64

s0.8m/s9.810.42

22

max

=

=π

A

115 ••• Picture the Problem In (b), we can use the condition Fnet = dU/dx = 0 for stable equilibrium to find the value of x = x0 at stable equilibrium. In (c) and (d), we can simply follow the outline provided in the problem statement. In (e), we can obtain the frequency

from mkf

π21

= using the value for k from the potential function.

(a) A graph of U(x) follows:

0123456789

10

0.0 0.5 1.0 1.5 2.0 2.5 3.0

x /a

U/U

0

(b) Express the condition for equilibrium: 0==dxdUF

Differentiate U with respect to x:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎥⎦

⎤⎢⎣⎡ −=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +=

2

20

20

0

11xa

aU

xa

aU

xa

axU

dxd

dxdU

Oscillations

1133

Set this derivative equal to zero and solve for x:

01 20

20 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−

xa

aU

and ax =0 or 1=α

(c) Express U(x0 + ε): ( )

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++=

⎥⎦

⎤⎢⎣

⎡+

++

=+

aaxaa

xU

xa

axUxU

εε

εεε

0

00

0

000

1

or, because x0 = a,

( )

( )[ ]10

00

11

1

11

−+++=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++=+

ββ

εεε

U

aa

UxU

whereaεβ =

(d) Expand ( ) 11 −+ β to obtain: ( ) ( ) ( )( )

2

21

1

...12

21111

ββ

βββ

+−≈

+×−−

+−+=+ −

Substitute in U(x0 + ε): ( ) [ ]

[ ]

2

2

0

2

2

00

20

200

constant

2

2

11

aU

aUU

U

UxU

ε

ε

β

βββε

+=

+=

+=

+−++=+

(e) Express the potential energy of a simple harmonic oscillator:

221constant εkU +=

If the particle whose potential energy is given in part (d) is to undergo simple harmonic motion:

202

aUk =

Chapter 14

1134

Express the frequency of the simple harmonic motion, substitute for k, and simplify to obtain:

mU

a

maU

mkf

0

20

221

221

21

π

ππ

=

==

116 ••• Picture the Problem Let m represent the mass of the cylindrical drum, R its radius, and k the stiffness constant of the spring. We can find the angular frequency of the oscillations by equating the maximum kinetic energy of the drum and the maximum energy stored in the spring. We can then express the frequency of the system in terms of its angular frequency. The application of Newton’s 2nd law, under on-the-verge-of-sliding conditions, together with the introduction of the oscillator’s total energy, will lead us to an expression for the minimum value of the coefficient of static friction. (a) Express the frequency of oscillation of the system for small displacements from equilibrium:

πω2

=f (1)

Express the kinetic energy of the drum and simplify to obtain:

( )2

43

221

22

21

21

2212

21

mv

mvRvmR

mvIK

=

+⎟⎠⎞

⎜⎝⎛=

+= ω

Apply conservation of energy to obtain:

2212

max43

max kAmvK ==

Substitute Aω for vmax: ( ) 2212

43 kAAm =ω

Solve for ω:

mk

32

=ω


mkf

32

21π

=


( )( ) Hz36.3

kg63N/m40002

21

==π

f

(b) Apply ∑ = 0xF to the drum to 0maxs, =− fkA

or

Oscillations

1135

establish the condition that governs slipping:

0ns =− FkA µ

Using Fn = mg, solve for µs: mgkA

=sµ (2)

Express the oscillator’s total energy in terms of the amplitude of its motion:

EkkAkAE 2221 =⇒=


mgEk2

s =µ

Substitute numerical values and evaluate µs:

( )( )( )( ) 40.3

m/s9.81kg6N/m4000J52

2s ==µ

*117 ••• Picture the Problem The pictorial representation shows the two blocks connected by the spring and displaced from their equilibrium positions. We can apply Newton’s 2nd law to each of these coupled oscillators and solve the resulting equations simultaneously to obtain the differential equation of motion of the coupled oscillators. We can then compare this differential equation and its solution to the differential equation of motion of the simple harmonic oscillator and its solution to show that the oscillation frequency is

( ) 21µω k= where µ = m1m2/(m1 + m2) is the reduced mass of the system.

Apply ∑ = aF rr

m to the block whose mass is m1 and solve for its acceleration:

( ) 21

2

11121 dtxdmamxxk ==−

or

( )211

21

2

1 xxmk

dtxda −==

Apply ∑ = aF rr

m to the block whose mass is m2 and solve for its

( ) 22

2

12221 dtxdmamxxk ==−−

Chapter 14

1136

acceleration:

or

( )122

22

2

2 xxmk

dtxda −==

Subtract the first equation from the second to obtain:

( ) xmm

kdt

xddt

xxd⎟⎟⎠

⎞⎜⎜⎝

⎛+−==

−

212

2

212

2 11

where x = x2 − x1

The reduced mass of the system is:

21

111mm

+=µ

or 21

21

mmmm+

=µ

Substitute to obtain: xk

dtxd

µ−=2

2

(1)

Compare this differential equation with the differential equation of the simple harmonic oscillator:

xmk

dtxd

−=2

2

The solution to this equation is: ( )δω += txx cos0

where mk

=ω

Express the solution to equation (1): ( )δω += txx cos0

where µ

ω k=

118 •• Picture the Problem We can use ( ) 21µω k= and µ = m1m2/(m1 + m2) from Problem 117 to find the spring constant for the HCl molecule. Use the result of Problem 118 to relate the oscillation frequency to the spring constant and reduced mass of the HCl molecule:

µω k=

Solve for k to obtain: 2µω=k

Express the reduced mass of the HCl molecule: 21

21

mmmm+

=µ


21

221

mmmm

k+

=ω

Oscillations

1137

Express the masses of the hydrogen and Cl atoms:

m1 = 1 amu = 1.67×10-27 kg and m2 = 35.45 amu = 5.92×10-26 kg


( )( )( ) N/m1.13kg1092.5kg1067.1

s10969.8kg1092.5kg1067.12627

21-132627

=×+×

×××= −−

−−

k

119 •• Picture the Problem In Problem 117, we derived an expression for the oscillation frequency of a spring-and-two-block system as a function of the stiffness constant of the spring and the reduced mass of the two blocks. We can solve this problem, assuming that the "spring constant" does not change, by using the result of Problem 117 and the reduced mass of a deuterium atom and a Cl atom in the equation for the oscillation frequency. Use the result of Problem 117 to relate the oscillation frequency to the spring constant and reduced mass of the HCl molecule:

µω k=

Express the reduced mass of the HCl molecule: 21

21

mmmm+

=µ

Express the masses of the deuterium and Cl atoms:

m1 = 2 amu = 3.34×10-27 kg and m2 = 35.45 amu = 5.92×10-26 kg

Evaluate the reduced mass of the molecule:

( )( )

kg1016.3kg1092.5kg1034.3kg1092.5kg1034.3

27

2627

2627

−

−−

−−

×=

×+×××

=µ


rad/s106.44

kg103.16N/m1.13

13

27-

×=

×=ω

120 ••• Picture the Problem The pictorial representation shows the block moving from right to left with an instantaneous displacement x from its equilibrium position. The free-body diagram shows the forces acting on the block during the half-cycles that it moves from right to left. When the block is moving from left to right, the directions of the kinetic friction force and the restoring force exerted by the spring are reversed. We can apply Newton’s 2nd law to these motions to obtain the differential equations given in the problem statements and then use their solutions to plot the graph called for in (c).

Chapter 14

1138

(a) Apply xx maF =∑ to the block while it is moving to the left to obtain:

2

2

k dtxdmkxf =−

Using fk = µkFn =µkmg, eliminate fk in the differential equation of motion:

mgkxdt

xdm k2

2

µ+−=

or

⎟⎠⎞

⎜⎝⎛ −−=

kmgxk

dtxdm k2

2 µ

Let kmgx k

0µ

= to obtain: ( )02

2

xxkdt

xdm −−=

or

x'x'mk

dtx'd 22

2

ω−=−=

provided x′ = x − x0 and

2kk

0 ωµµ g

kmgx ==

The solution to the differential equation is:

( )δω += t'xx' cos0 and its derivative is

( )δωω +−= t'xv' sin0

The initial conditions are:

( ) 00 xxx' −= and ( ) 00 =v'

Apply these conditions to obtain: ( ) 00 cos0 xx'xx' −== δ and ( ) 0sin0 0 =−= δω 'xv'


0=δ and 00 xx'x −= and

( ) txxx' ωcos0−=

Oscillations

1139

or ( ) 00 cos xtxxx +−= ω (1)

(b) Apply aF rr

m=∑ to the block while it is moving to the right to obtain:

2

2

k dtxdmkxf =−−

Using fk = µkFn =µkmg, eliminate fk in the differential equation of motion:

mgkxdt

xdm k2

2

µ−−=

or

⎟⎠⎞

⎜⎝⎛ +−=

kmgxk

dtxdm k2

2 µ

Let kmgx k

0µ

= to obtain: ( )02

2

xxkdt

xdm +−=

or

x"x"mk

dtx"d 22

2

ω−=−=

provided x″ = x + x0 and

2kk

0 ωµµ g

kmgx == .

The solution to the differential equation is:

( )δω += t"xx" cos0 and its derivative is

( )δωω +−= t"xv" sin0

The initial conditions are:

( ) 00 xxx" += and ( ) 00 =v"

Apply these conditions to obtain: ( ) 00 cos0 xx"xx" +== δ and ( ) 0sin0 0 =−= δω "xv"


0=δ and 00 xx"x += and

( ) txxx" ωcos0+= or ( ) 00 cos xtxxx −+= ω (2)

(c) A spreadsheet program to calculate the position of the oscillator as a function of time (equations (1) and (2)) is shown below. The constants used in the position functions (x0 = 1 m and T = 2 s were used for simplicity) and the formulas used to calculate the positions are shown in the table. After each half-period, one must compute a new amplitude for the oscillation, using the final value of the position from the last half-period.

Chapter 14

1140

Cell Content/Formula Algebraic Form B1 1 x0 B2 10 A C7 C6 + 0.1 t + ∆t D7 ($B$2−$B$1)*COS(PI()*C7)+$B$1 ( ) 00 cos xtxA +− π

D17 (ABS($D$6+$B$1))*COS(PI()*C17)−$B$1 00 cos xtxx −+ π

D27 (ABS($D$6−$B$1))*COS(PI()*C27)+$B$1 00 cos xtxx +− π

D37 (ABS($D$36+$B$1))*COS(PI()*C37)−$B$100 cos xtxx −+ π

D47 ($D$46−$B$1)*COS(PI()*C47)+$B$1 ( ) 00 cos xtxx +− π

A B C D 1 x0= 1 m 2 A= 10 3 4 t x 5 (s) (m) 6 0.0 10.007 0.1 9.56 8 0.2 8.28 9 0.3 6.29 10 0.4 3.78

53 4.7 0.41 54 4.8 0.19 55 4.9 0.05 56 5.0 0.00

The graph shown below was plotted using the data from columns C (t) and D (x). cycles.-half fiveafter ceasesblock theofmotion that theNote

-10

-8

-6

-4

-2

0

2

4

6

8

10

0 1 2 3 4 5

t (s)

x (m

)

Oscillations

1141

121 ••• Picture the Problem The diagram shows the half-cylinder displaced from its equilibrium position through an angle θ. The frequency of its motion will be found by expressing the mechanical energy E in terms of θ and dθ/dt. For small θ we will obtain an equation of

the form .2

212

21 ⎟

⎠⎞

⎜⎝⎛+=

dtdIE θκθ Differentiating both sides of this equation with respect

to time will lead to dtd

dtdI θθκθ ⎟⎟

⎠

⎞⎜⎜⎝

⎛+= 2

2

0 , an equation that must be valid at all times.

Because the situation of interest to us requires that dθ/dt is not always equal to zero, we

have 2

2

0dtdI θκθ += or 02

2

=+ θκθIdt

d, the differential equation of simple harmonic

motion with .2 Iκω = The distance from O to the center of mass D, where, from

Problem 8-39, D = (4/3π)R, and the distance from the contact point C to the center of mass is r. Finally, we’ll take the potential energy to be zero where θ is zero and assume that there is no slipping.


( )2

21

⎟⎠⎞

⎜⎝⎛+−=

+=

dtdIDhMg

KUE

Cθ

(1)

From Table 9-1, the moment of inertia of a solid cylinder about an axis perpendicular to its face and through its center is given by:

( ) 22cylinder solid,0 2

21 MRRMI ==

where M is the mass of the half-cylinder.

Chapter 14

1142

Express the moment of inertia of the half-cylinder about the same axis:

[ ] 220cylinderhalf 0, 2

121 MRMRII ===

Use the parallel-axis theorem to relate Icm to I0:

2cm0 MDII +=

Substitute for Icm and solve for Icm:

MDMR

MDII

22

20cm

21

−=

−=

Apply the parallel-axis theorem a second time to obtain an expression for IC: ⎟

⎠⎞

⎜⎝⎛ +−=

+−=

222

222C

21

21

rDRM

MrMDMRI (2)

Apply the law of cosines to obtain:

θcos2222 RDDRr −+=

Substitute for r2 in equation (2) to obtain:

⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −++−= θθ cos2

23cos2

21 22222

C RDMRRDDRDRMI

Substitute for h and IC in equation (1):

( )2

2 cos223

21cos1 ⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛ −+−=

dtd

RDMRMgDE θθθ

Use the small angle approximation 2

211cos θθ −≈ to obtain:

[ ]2

222 223

21

21

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛ −−+=

dtd

RDMRMgDE θθθ

Because θ 2 << 2, we can neglect the θ 2 in the square brackets to obtain:

222 2

23

21

21

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛ −+=

dtd

RDMRMgDE θθ

Oscillations

1143

Differentiating both sides with respect to time yields:

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛ −+= 2

22 2

230

dtd

dtd

RDMR

dtdMgD θθθθ ,

0223

2

22 =+⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ − θθ gD

dtd

RDR ,

and

02

232

2

2

=⎟⎠⎞

⎜⎝⎛ −

+ θθ

RDR

gDdtd

,

the differential equation of simple harmonic motion with ⎟⎠⎞

⎜⎝⎛ −

=

RDR

gD

2232

2ω .

Substitute for D to obtain:

Rg

Rg

⎟⎠⎞

⎜⎝⎛

−=

⎟⎠⎞

⎜⎝⎛ −

=169

8

38

23

34

2

ππ

πω

Express the period of the motion in terms of ω and simplify to obtain:

gR

gRT

78.7

816922

=

⎟⎠⎞

⎜⎝⎛ −

==ππ

ωπ

*122 ••• Picture the Problem The net force acting on the particle as it moves in the tunnel is the x-component of the gravitational force acting on it. We can find the period of the particle from the angular frequency of its motion. We can apply Newton’s 2nd law to the particle in order to express ω in terms of the radius of the earth and the acceleration due to gravity at the surface of the earth. (a) From the figure we see that:

xR

GmM

rxr

RGmMFF rx

3E

E

3E

Esin

−=

−== θ

Because this force is a linear restoring

force, the motion of the particle is simple

Chapter 14

1144

harmonic motion.

(b) Express the period of the particle as a function of its angular frequency:

ωπ2

=T (1)

Apply ∑ = xx maF to the particle: maxR

GmM=− 3

E

E

Solve for a: xx

RGMa 2

3E

E ω−=−=

where

2E

E

RGM

=ω

Use 2

EE gRGM = to simplify ω:

E3E

2E

Rg

RgR

==ω


gR

Rg

T E

E

22 ππ==


min4.84

s1006.5m/s9.81

m106.372 32

6

=

×=×

= πT

123 ••• Picture the Problem The amplitude of a damped oscillator decays with time according

to ( ) .20

tmbeAA −= We can find b/2m from 2

00 2

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

ωωω

mb' and then substitute in

the amplitude equation to find the factor by which the amplitude is decreased during each oscillation. We’ll use our result from (a), together with the dependence of the energy of the oscillator on the square of its amplitude, to find the factor by which its energy is reduced during each oscillation. (a) Express the variation in amplitude with time:

( )tmbeAA 20

−= (1)

Oscillations

1145

Relate the damped and undamped frequencies of the oscillator:

2

00 2

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

ωωω

mb' (14-46)

Solve for mb 2 :

( )

0

202

0

2

0

436.0

9.01'12

ω

ωωωω

=

−=−=mb

Find the period of the damped oscillations: 09.0

22ωπ

ωπ==

'T

Substitute in equation (1) with t = T to obtain:

( )

0478.0

04.39.02436.0

0

00

=

== −⎟⎟⎠

⎞⎜⎜⎝

⎛−

eeAA ω

πω

(b) Express the energy of the oscillator at time t = 0:

202

10 kAE =

Express the energy of the oscillator at time t = T:

221 kAE =

Divide the second of these equations by the first, simplify, and substitute to evaluate E/E0:

( )

00228.0

0477.0 22

020

2

0

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛==

AA

AA

EE

124 ••• Picture the Problem We can differentiate Equation 14-52 twice and substitute x and d2x/dt2 in Equation 14-51 to determine the condition that must be satisfied in order for Equation 14-52 to be a solution of Equation 14-51. The differential equation of motion is Equation 14-51:

tFxmdtdxb

dtxdm ωω cos0

202

2

=++

Its proposed solution is Equation 14-52:

( )δω −= tAx cos

Obtain the first and second derivatives of x:

( )δωω −−= tAdtdx sin

and

Chapter 14

1146

( )δωω −−= tAdt

xd cos22

2

Substitute in the differential equation to obtain:

( ) ( ) ( ) tFtAmtbAtmA ωδωωδωωδωω coscossincos 020

2 =−+−−−−

Using trigonometric identities, expand ( )δω −tcos and ( )δω −tsin to obtain:

( ) ( )

( ) tFttAmttbAttmA

ωδωδωω

δωδωωδωδωω

cossinsincoscossincoscossinsinsincoscos

020

2

=++

−−+−

Factor ( )δωδω sinsincoscos ttmA + from the first and third terms to obtain:

( )( ) ( ) tFttbAttmA ωδωδωωδωδωωω cossincoscossinsinsincoscos 0

220 =−−+−

Factor δω coscos t from the first term on the left-hand side of the equation and

δω cossin t from the 2nd term:

( )( ) ( )

tFtttbA

tttmA

ωδωδωδωω

δωδωδωωω

coscossinsincos1cossin

coscossinsin1coscos

0

220

=

⎟⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ +−

Simplify to obtain:

( )( )( ) ( )

tFt

tbAttmA

ωωδδωωδωδωωω

costantan1cossintantan1coscos

0

220

=

⎟⎠⎞

⎜⎝⎛ −−+−

Divide both sides of the equation by ( )22

0 ωω −m :

( )( ) ( )( )

( ) tm

Ft

tm

bAttA

ωωω

ωδδω

ωωωδωδω

cos

tantan1cossintantan1coscos

220

0

220

−=

⎟⎠⎞

⎜⎝⎛ −

−−+

The phase constant for a driven oscillator is given by Equation 14-54:

( )220

tanωω

ωδ−

=m

b

Oscillations

1147

Substitute for δtan :

( ) ( ) ( )( )

( )( ) t

mF

tm

b

tm

bAm

bttA

ωωωω

ωωω

δωωω

ωωω

ωωδω

costan

1

cossintan1coscos

220

022

0

220

220

−=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛−

−×

−−⎟⎟⎠

⎞⎜⎜⎝

⎛−

+

Simplify to obtain:

( )( ) ( ) tm

FtA ωωω

δδω costan1coscos 220

02

−=+

Use the trigonometric identity δ

δ 22

cos1tan1 =+ :

( ) ( ) tm

FtA ωωωδ

δω coscos

1coscos 220

02 −

=

Simplify to obtain:

( ) tm

FtA ωωωδω coscoscos 22

0

0

−=

Thus ( )δω −= tAx cos is a

solution to Equation 14-51 provided:

( )220

0 cosωωδ

−=

mFA

*125 ••• Picture the Problem We can follow the step-by-step instructions provided in the problem statement to obtain the desired results. (a) Express the average power delivered by a driving force to a driven oscillator:

θcosFvP =⋅= vF rr

or, because θ is 0°, FvP =

Express F as a function of time: tFF ωcos0=

Express the position of the driven oscillator as a function of time:

( )δω −= tAx cos

Chapter 14

1148

Differentiate this expression with respect to time to express the velocity of the oscillator as a function of time:

( )δωω −−= tAv sin

Substitute to express the average power delivered to the driven oscillator:

( ) ( )[ ]( )δωωω

δωωω

−−=

−−=

ttFA

tAtFP

sincos

sincos

0

0

(b) Expand ( )δω −tsin to obtain: ( ) δωδωδω sincoscossinsin ttt −=−

Substitute in your result from (a) and simplify to obtain:

()

ttFAtFA

tttFAP

ωωδωωδω

δωδωωω

sincoscoscossin

sincoscossincos

0

20

0

−=

−−=

(c) Integrate θθ cossin over one period to determine θθ cossin :

0

sin21

21

cossin21cossin

2

0

2

2

0

=⎥⎥⎦

⎤

⎢⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡= ∫

π

π

θπ

θθθπ

θθ d

Integrate θ2cos over one period to determine θ2cos :

( )

( )210

21

2cos21

2cos121

21

cos21cos

2

021

2

021

2

0

2

0

22

=+=

⎥⎦

⎤⎢⎣

⎡+=

⎥⎦

⎤⎢⎣

⎡+=

=

∫∫

∫

∫

ππ

θθθπ

θθπ

θθπ

θ

ππ

π

π

dd

d

d

Substitute and simplify to express Pav:

( )δω

δωδωωωδω

ωδω

sin

0cossinsincoscos

cossin

021

0021

0

20av

FA

FAFAttFA

tFAP

=

−=

−

=

Oscillations

1149

(d) Construct a triangle that is consistent with

( )220

tanωω

ωδ−

=m

b:

Using the triangle, express sinδ:

( ) 222220

2sin

ωωω

ωδbm

b

+−=

Using equation 14-53, reduce this expression to the simpler form:

0

sinF

Abωδ =

(e) Solve 0

sinF

Abωδ = for ω: δω sin0

bAF

=

Substitute in the expression for Pav to eliminate ω:

δ22

0av sin

2bFP =

Substitute for δsin from (d) to obtain Equation 14-55: ( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡

+−=

222220

2

20

2

av 21

ωωωω

bmFbP

126 ••• Picture the Problem We can follow the step-by-step instructions given in the problem statement to derive the given results. (a) Express the condition on the denominator of Equation 14-55 when the power input is half its maximum value:

( ) 20

2222220

2 2 ωωωω bbm =+−

and, for a sharp resonance,

( ) 20

22220

2 ωωω bm ≈−

Factor the difference of two squares to obtain:

( )( )[ ] 20

2200

2 ωωωωω bm ≈+−

or

( ) ( ) 20

220

20

2 ωωωωω bm ≈+−

(b) Use the approximation ω + ω0 ≈ 2ω0 to obtain:

( ) ( ) 20

220

20

2 2 ωωωω bm ≈−

Chapter 14

1150

Solve for ω0 − ω: mb

20 ±=−ωω (1)

(c) Using its definition, express Q:

bmQ 0ω=

Solve for b:

Qmb 0ω=

(d) Substitute for b in equation (1) to obtain: Q2

00

ωωω ±=−

Solve for ω:

Q20

0ωωω ±=

Express the two values of ω:

Q20

0ωωω +=+

and

Q20

0ωωω −=−

Remarks: Note that the width of the resonance at half-power is ,∆ Qωωωω 0=−= −+ in agreement with Equation 14-49. 127 ••• Picture the Problem We can find the equilibrium separation for the Morse potential by setting dU/dr = 0 and solving for r. The second derivative of U will give the "spring constant" for small displacements from equilibrium. In (c), we can use ,µω k= where k is our result from (b) and µ is the reduced mass of a homonuclear diatomic molecule, to find the oscillation frequency of the molecule. (a) A spreadsheet program to calculate the Morse potential as a function of r is shown below. The constants and cell formulas used to calculate the potential are shown in the table.

Cell Content/Formula Algebraic Form B1 5 D B2 0.2 β C9 C8 + 0.1 r + ∆r D8 $B$1*(1−EXP(−$B$2*(C8−$B$3)))^2 ( )[ ]2

01 rreD −−− β

Oscillations

1151

A B C D

1 D= 5 eV 2 Beta= 0.2 nm−1 3 r0= 0.75 nm 4 5 6 r U(r) 7 (nm) (eV) 8 0.0 0.130959 0.1 0.09637

10 0.2 0.0676011 0.3 0.0443412 0.4 0.02629

235 22.7 4.87676236 22.8 4.87919237 22.9 4.88156238 23.0 4.88390239 23.1 4.88618

The graph shown below was plotted using the data from columns C (r) and D (U(r)).

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.0 0.5 1.0 1.5 2.0 2.5 3.0

r (nm)

U (e

V)

(b) Differentiate the Morse potential with respect to r to obtain:

( )[ ] [ ])(

2

0

0

12

1

rr

rr

eD

eDdrd

drdU

−−

−−

−−=

−=

β

β

β

Set this derivative equal to zero for extrema:

[ ] 012 )( 0 =−− −− rreD ββ

Chapter 14

1152

Solve for r to obtain: 0rr =

Evaluate the second derivative of U(r) to obtain:

[ ] )(2

)(2

2

0

0

2

12

rr

rr

De

eDdrd

drUd

−−

−−

=

−−=

β

β

β

β

Evaluate this derivative at r = r0: D

drUd

rr

22

2

20

β==

(1)

Recall that the potential function for a simple harmonic oscillator is:

221 kxU =

Differentiate this expression twice to obtain:

kdx

Ud=2

2

By comparison with equation (1) we have:

Dk 22β=

(c) Express the oscillation frequency of the diatomic molecule:

µω k=

where µ is the reduced mass of the molecule.

Express the reduced mass of the homonuclear diatomic molecule:

22

2

21

21 mm

mmm

mm==

+=µ


mD

mD ββω 2

2

2 2

==

Remarks: An alternative approach in (b) is the expand the Morse potential in a Taylor series

( ) ( ) ( ) ( ) ( ) ( ) termsorder higher 2!1 r'U'rrrU'rrrUrU 0

20000 +−+−+=

to obtain ( )202 rrDβU(r) −≈ . Comparing this expression to the energy of a spring-

and-mass oscillator we see that, as was obtained above, Dβk 22= .

1153

Chapter 15 Wave Motion Conceptual Problems *1 • Determine the Concept The speed of a transverse wave on a rope is given by

µFv = where F is the tension in the rope and µ is its linear density. The waves on the

rope move faster as they move up because the tension increases due to the weight of the rope below. 2 • Determine the Concept The distance between successive crests is one wavelength and the time between successive crests is the period of the wave motion. Thus, T = 0.2 s and f = 1/T = 5 Hz. correct. is )(b

3 • Picture the Problem True. The energy per unit volume (the average energy density) is given by 2

02

21

av sρωη = where s0 is the displacement amplitude.

4 • Determine the Concept At every point along the rope the wavelength, speed, and frequency of the wave are related by .fv=λ The speed of the wave, in turn, is given

by µFv = where F is the tension in the rope and µ is its linear density. Due to the

weight of the rope below, the tension is greater at the top and the speed of the wave is also greater at the top. Because v,∝λ the wavelength is greater at the top. *5 • Determine the Concept The speed of the wave v on the bullwhip varies with the tension F in the whip and its linear density µ according to µFv = . As the whip tapers, the

wave speed in the tapered end increases due to the decrease in the mass density, so the wave travels faster. 6 • Picture the Problem The intensity level of a sound wave β, measured in decibels, is given by ( ) ( )0logdB10 II=β where I0 = 10-12 W/m2 is defined to be the threshold of

hearing. Express the intensity of the 60-dB sound:

( )0

60logdB10dB60II

= ⇒ 06

60 10 II =

Chapter 15

1154

Express the intensity of the 30-dB sound:

( )0

30logdB10dB30II

= ⇒ 03

30 10 II =

Because I60 = 103I30: false. isstatement The

7 • Determine the Concept No. Because the source and receiver are at rest relative to each other, there is no relative motion of the source and receiver and there will be no Doppler shift in frequency. 8 • Determine the Concept Because there is no relative motion of the source and receiver, there will be no Doppler shift and the observer will hear sound of frequency f0.

correct. is )(a

*9 •• Determine the Concept The light from the companion star will be shifted about its mean frequency periodically due to the relative approach to and recession from the earth of the companion star as it revolves about the black hole. 10 • Determine the Concept In any medium, the wavelength, frequency, and speed of a wave are related through λ = v/f. Because the frequency of a wave is determined by its source and is independent of the nature of the medium, if v differs in the two media, the wavelengths will also differ. In this situation, the frequencies are the same but the speeds of propagation are different. 11 • (a) True. The particles that make up the string move at right angles to the direction the wave is propagating. (b) False. Sound waves in air are longitudinal waves of compression and rarefaction. (c) False. The speed of sound in air varies with the square root of the absolute temperature. The speed of sound at 20°C is 4% greater than that at 5°C. 12 • Determine the Concept In any medium, the wavelength, frequency, and speed of a sound wave are related through λ = v/f. Because the frequency of a wave is determined by its source and is independent of the nature of the medium, if v is greater in water than

Wave Motion

1155

in air, λ will be greater in water than in air. correct. is )(a

*13 • Determine the Concept There was only one explosion. Sound travels faster in water than air. Abel heard the sound wave in the water first, then, surfacing, heard the sound wave traveling through the air, which took longer to reach him. 14 •• Determine the Concept The graph shown below shows the pulse at an earlier time (−∆t) and later time (∆t). One can see that at t = 0, the portion of the string between 1 cm and 2 cm is moving down, the portion between 2 cm and 3 cm is moving up, and the string at x = 2 cm is instantaneously at rest.

0 1 2 3 4 5 6 7 8 9 10

x , cm

y

t = 0t > 0

t < 0

15 •• Determine the Concept The velocity of the string at t = 0 is shown. Note that the velocity is negative for 1 cm < x < 2 cm and is positive for 2 cm < x < 3 cm.

0 1 2 3 4 5 6 7 8 9 10

x , cm

v y

Chapter 15

1156

16 •• Determine the Concept As the jar is evacuated, the speed of sound inside the jar decreases. Because of the mismatch between the speed of sound inside and outside of the jar, a larger fraction of the sound wave is reflected back into the jar, and a smaller fraction is transmitted through the glass of the bell jar. *17 •• Determine the Concept Path C. Because the wave speed is highest in the water, and more of path C is underwater than A or B, the sound wave will spend the least time on path C. Estimation and Approximation 18 •• Picture the Problem The rate at which energy is delivered by sound waves is the product of its intensity and the area over which the energy is delivered. We can use the definition of the intensity level of the speech at 1 m to find the intensity of the sound and the formula for the area of a sphere to find the area over which the energy is distributed. Express the power of human speech as a function of its intensity:

IAP =

Express the area of a sphere of radius 1 m:

( ) 222 m4m144 πππ === rA

Use ( ) ( )0logdB10 II=β to solve

for the intensity of the sound at the 65-dB level:

( )0

logdB10dB65II

=

and ( )

26

2125.60

5.6

W/m1016.3

W/m101010−

−

×=

== II

Substitute and evaluate P: ( )( )

W1097.3

m4W/m1016.35

226

−

−

×=

×= πP

19 •• Picture the Problem Let d represent the distance from the bridge to the water under the assumption that the time for the sound to reach the man is negligible; let t be the elapsed time between dropping the stone and hearing the splash. We’ll use a constant-acceleration equation to find the distance to the water in all three parts of the problem, just improving our initial value with corrections taking into account the time required for the sound of the splash to reach the man on the bridge.

Wave Motion

1157

(a) Using a constant-acceleration equation, relate the distance the stone falls to its time-of-fall:

221

0 gttvd +=

or, because v0 = 0, 2

21 gtd =


( )( ) m78.5s4m/s9.81 2221 ==d

(b) Express the actual distance to the water d′ in terms of a time correction ∆t:

( )221 ttgd' ∆−=

Express ∆t:

svdt =∆


2

s21' ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

vdtgd

Substitute numerical values and evaluate d′: ( )

m69.7

m/s340m78.5s4m/s9.81

22

21

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=d'

(c) Express the total time for the rock to fall and the sound to return to the man:

ssoundrockfalling g

2vddttt +=∆+∆=∆

Rewrite the equation in quadratic form: ( ) 012 22

ss

2s

2 =∆+⎟⎟⎠

⎞⎜⎜⎝

⎛ ∆+− tvd

vt

gvd


( ) 0m1085.1m1063.2 2642 =×+×− dd

Solve for the positive value of d: m5.70=d … about 1% larger than

our result in part (b) and 11% smaller than

our first approximation in (a).

*20 •• Picture the Problem You can use a protractor to measure the angle of the shock cone and then estimate the speed of the bullet using .sin uv=θ The speed of sound in helium at room temperature (293 K) is 977 m/s.

Chapter 15

1158

Relate the speed of the bullet u to the speed of sound v in helium and the angle of the shock cone θ :

uv

=θsin

Solve for u: θsin

vu =

Measure θ to obtain:

°≈ 70θ

Substitute numerical values and evaluate u:

km/s04.170sinm/s977

=°

=u

21 •• Picture the Problem Let d be the distance to the townhouses. We can relate the speed of sound to the distance to the townhouses to the frequency of the clapping for which no echo is heard. Relate the speed of sound to the distance it travels to the townhouses and back to the elapsed time:

tdv∆

=2

Express d in terms of the number of strides and distance covered per stride:

( )( ) m54m/stride8.1strides30 ==d

Relate the elapsed time ∆t to the frequency f of the clapping:

s4.0claps/s5.211

===∆f

t


( ) m/s270s4.0m542

==v

Express the percent difference between this result and 340 m/s: %6.20

m/s340m/s270m/s340

=−

Speed of Waves

22 • Picture the Problem The speed of sound in a fluid is given by ρBv = where B is the

bulk modulus of the fluid and ρ is its density. (a) Express the speed of sound in water in terms of its bulk modulus:

ρBv =

Substitute numerical values and evaluate v: km/s1.41

kg/m10N/m102.0

33

29

=×

=v

Wave Motion

1159

(b) Solve ρBv = for B: 2vB ρ=

Substitute numerical values and evaluate B:

( )( )210

233

N/m102.70

m/s1410kg/m1013.6

×=

×=B

*23 • Picture the Problem The speed of sound in a gas is given by MRTv γ= where R is

the gas constant, T is the absolute temperature, M is the molecular mass of the gas, and γ is a constant that is characteristic of the particular molecular structure of the gas. Because hydrogen gas is diatomic, γ = 1.4. Express the dependence of the speed of sound in hydrogen gas on the absolute temperature:

MRTv γ

=


( )( )

km/s32.1

kg/mol102K300KJ/mol314.84.1

3

=

×⋅

= −v

24 • Picture the Problem The speed of a transverse wave pulse on a wire is given by

µFv = where F is the tension in the wire, m is its mass, L is its length, and µ is its

mass per unit length. Express the dependence of the speed of the pulse on the tension in the wire:

µFv =

where µ is the mass per unit length of the wire.

Substitute numerical values and evaluate v: m/s251

m7kg0.1N900

===Lm

Fv

25 • Picture the Problem The speed of transverse waves on a wire is given by


mass per unit length.

Chapter 15

1160

Express the dependence of the speed of the pulse on the tension in the wire and the linear density of the wire:

LmFv =

Solve for m: 2v

FLm =


( )( )( )

g19.6m/s150

m0.8N5502 ==m

*26 • Picture the Problem The speed of a wave pulse on a wire is given by µFv = where

F is the tension in the wire, m is its mass, L is its length, and µ is its mass per unit length. (a) Doubling the length while keeping the mass per unit length constant does not change the linear density:

m/s20=v

(b) Because v depends of ,F

doubling the tension increases v by a factor of 2 :

( ) m/s28.3m/s202 ==v

(c) Because v depends on ,1 µ

doubling µ reduces v by a factor of 2 :

m/s1.412m/s20

==v

27 • Picture the Problem The speed of a transverse wave on the piano wire is given by


mass per unit length. (a) The speed of transverse waves on the wire is given by:

LmFFv ==

µ

Substitute numerical values and evaluate v: ( ) ( ) m/s265

m0.7kg0.005N500

==v

Wave Motion

1161

(b) Letting m′ represent the mass of the wire when copper has been wrapped around the steel wire, express ∆m, the amount of copper wire required:

mm'm −=∆

Express the new wave speed v′: Lm'

Fv' =

Express the ratio of the speed of the waves in part (a) to the reduced wave speed: m

m'

Lm'F

LmF

v'v

=== 2

Solve for and evaluate m′:

mm' 4=

Substitute to obtain: ( )g15.0

g534

=

=−=−=∆ mmmm'm

28 •• Picture the Problem We can estimate the accuracy of this procedure by comparing the estimated distance to the actual distance. Whether a correction for the time it takes the light to reach you is important can be decided by comparing the times required for light and sound to travel a given distance. (a) Convert 340 m/s to km/s: km/s340.0

m1000km1

sm034 =×=v

(b) Express the fractional error in the procedure:

%06.2340.0007.0

340.0333.0340.0estimated

==

−=

−=

∆t

tts

ssss

(c) Compare the time required for light to travel 1 km to the time required for sound to travel the same distance:

68

sound

light 10m/s103

m/s340km1

km1−≈

×===

∆∆

cv

v

ctt

Chapter 15

1162

important.not isyou reach light tofor timefor the correction a small, so isfraction thisBecause

*29 •• Picture the Problem The speed of a transverse wave on a string is given by

µFv = where F is the tension in the wire and µ is its linear density. We can

differentiate this expression with respect to F and then separate the variables to show that the differentials satisfy .2

1 FdFvdv = We’ll approximate the differential quantities to

determine by how much the tension must be changed to increase the speed of the wave to 312 m/s. (a) Evaluate dv/dF:

Fv

FF

dFd

dFdv

⋅==⎥⎦

⎤⎢⎣

⎡=

211

21

µµ

Separate the variables to obtain:

FdF

vdv

21

=

(b) Solve for dF:

vdvFdF 2=

Approximate dF with ∆F and dv with ∆v to obtain:

vvFF ∆

=∆ 2

Substitute numerical values and evaluate ∆F:

( ) N0.40m/s300

m/s12N5002 ==∆F

30 •• Picture the Problem The speed of sound in a gas is given by MRTv γ= where R is

the gas constant, T is the absolute temperature, M is the molecular mass of the gas, and γ is a constant that is characteristic of the particular molecular structure of the gas. We can differentiate this expression with respect to T and then separate the variables to show that the differentials satisfy .2

1 TdTvdv = We’ll approximate the differential quantities to

determine the percentage change in the velocity of sound when the temperature increases from 0 to 27°C. Lacking information regarding the nature of the gas, we can express the ratio of the speeds of sound at 300 K and 273 K to obtain an expression that involves just the temperatures.

Wave Motion

1163

(a) Evaluate dv/dT:

Tv

MR

RTM

MRT

dTd

dTdv

21

21

=

⎟⎠⎞

⎜⎝⎛=⎥

⎦

⎤⎢⎣

⎡=

γγ

γ


TdT

vdv

21

=

(b) Approximate dT with ∆T and dv with ∆v and substitute numerical values to obtain:

%95.4273K27

21

21

=⎟⎠⎞

⎜⎝⎛=

∆=

∆ KTT

vv

(c) Using a differential approximation, approximate the speed of sound at 300 K:

⎟⎠⎞

⎜⎝⎛ ∆+=

∆+≈

vvv

vvvvv

1K273

K273K273K300

Substitute numerical values and evaluate v300 K:

( )( ) m/s3470495.01m/s331K300 =+=v

Use MRTv γ= to express the

speed of sound at 300 K:

( )M

Rv K300K300

γ=

Use MRTv γ= to express the

speed of sound at 273 K:

( )M

Rv K273K732

γ=

Divide the first of these equations by the second and solve for and evaluate v300 K:

( )

( ) 273300

K273

K300

K273

K300 ==

MR

MR

vv

γ

γ

and

( ) m/s347273300m/s331K300 ==v

Note that these two results agree to three significant figures.

31 ••• Picture the Problem The speed of sound in a gas is given by MRTv γ= where R is

the gas constant, T is the absolute temperature, M is the molecular mass of the gas, and γ

Chapter 15

1164

is a constant that is characteristic of the particular molecular structure of the gas. Because T = t + 273 K, we can differentiate v with respect to t to show that ( ).2

1 Tvdtdv =

Evaluate dv/dt: ( )

( )

Tv

MR

tRM

MtR

dtd

dtdv

21

K27321

K273

=

⎟⎠⎞

⎜⎝⎛

+=

⎥⎦

⎤⎢⎣

⎡ +=

γγ

γ

Substitute for t to obtain:

⎥⎦

⎤⎢⎣

⎡+

=K2732

1t

vdtdv

Use the approximation

( ) ( ) TdTdvTvTv

T

∆⎟⎠⎞

⎜⎝⎛+≈

0

0

to write:

( ) ( )

( ) vCv

tt

vCvtv

∆+°=

⎥⎦

⎤⎢⎣

⎡+

+°≈

0K2732

10 (1)

where

tt

v ⎥⎦

⎤⎢⎣

⎡+

=∆K273

m/s33121

For t << 273 K: ( )ttv Km/s606.0

K273m/s331

21

⋅=⎥⎦

⎤⎢⎣

⎡≈∆


( )ttCvtv

Km/s606.0m/s331

Km/s606.00

⋅+=

⋅+°=

32 •• Picture the Problem Let d be the distance to the munitions plant, v1 be the speed of sound in air, v2 be the speed of sound in rock, and ∆t be the difference in the arrival times of the sound at the man’s apartment. We can express ∆t in terms of t1 and t2 and then express these travel times in terms of the distance d and the speeds of the sound waves in air and in rock to obtain an equation we can solve for the distance from the man’s apartment to the munitions plant. Express the difference in travel times for the sound wave transmitted through air and the sound wave transmitted through the earth:

ttt ∆=− 21

Wave Motion

1165

Express the transmission times in terms of the distance traveled and the speeds in the two media:

11 v

dt = and 2

2 vdt =


tvd

vd

∆=−21

or

tvv

vvdvv

d ∆=⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

21

12

21

11

Solve for d: t

vvvvd ∆−

=12

21


( )( ) ( )

km15.1

s3m/s340m/s3000m/s3000m/s340

=

−=d

33 ••• Picture the Problem The locations of the lightning strike (L), dorm room (R), and baseball park (P) are indicated on the diagram. We can neglect the time required for the electromagnetic pulse to reach the source of the radio transmission, which is the ballpark. The angle θ is related to the sides of the triangle through the law of cosines. We’re given the distance dPR and can find the other sides of the triangle using the speed of sound and the elapsed times.

Using the law of cosines, relate the angle θ to the distances that make up the sides of the triangle:

( ) θθ cos2180cos2 PRLP2PR

2LPPRLP

2PR

2LP

2LR ddddddddd ++=−°−+=

Solve for θ:

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−= −

PRLP

2PR

2LP

2LR1

2cos

dddddθ

Chapter 15

1166

Express the distance from the lightning strike to the ball park:

( )( ) m680s2m/s340LPsLP ==∆= tvd

Express the distance from the lightning strike to the dorm room:

( )( ) m2040s6m/s340LRsLR ==∆= tvd

Substitute and evaluate θ:

( ) ( ) ( )( )( ) °±=⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−= − 4.58

m1600m6802m1600m680m2040cos

2221θ

north. of E)(or W 58.4 ballpark, thefrom m 680struck lightning The °

*34 ••• Picture the Problem Choose a coordinate system in which station Alpha is at the origin and the axes are oriented as shown in the pictorial representation. Because 0.75 mi = 1.21 km, Alpha’s coordinates are (0, 0), Beta’s are (1.21 km, 0), and those of the lightning strike are (x, y). We can relate the distances from the stations to the speed of sound in air and the times required to hear the thunder at the two stations.

Relate the distance dα to the position coordinates of Alpha and the lightning strike:

222αdyx =+ (1)

Relate the distance dβ to the position coordinates of Beta and the lightning strike:

( ) 222km21.1 βdyx =+− (2)

Relate the distance dα to the speed of sound in air v and the time that elapses between seeing the lightning at Alpha and hearing the thunder:

( )( ) m1156s4.3m/s340 ==∆= αα tvd

Relate the distance dβ to the speed of sound in air v and the time that elapses between seeing the lightning at Beta and hearing the thunder:

( )( ) m850s5.2m/s340 ==∆= ββ tvd

Substitute in equations (1) and (2) to obtain:

( ) 2222 km336.1m1156 ==+ yx (3) and

Wave Motion

1167

( ) ( )2

222

km7225.0m850km21.1

=

=+− yx (4)

Subtract equation (4) from equation (3) to obtain:

( )2

222

km7225.0km336.1km21.1

−

=−− xx

or ( ) ( ) 22 km6135.0km21.1km42.2 =−x

Solve for x to obtain: km855.0=x


( ) 222 km336.1km855.0 =+ y

Solve for y, keeping the positive root because the lightning strike is to the north of the stations, to obtain:

km778.0=y

The coordinates of the lightning strike are:

( )km0.778 km,855.0

or ( )mi0.484 mi,531.0

35 ••• Picture the Problem The velocity of longitudinal waves on the Slinky is given by

ρBv = where B is the bulk modulus of the material from which the slinky is

constructed and ρ is its mass per unit volume. The velocity of transverse waves on the slinky is given by µFv = where F is the tension in the slinky and µ is its mass per

unit length. Substitution for B and ρ will lead to mkLv = in (a) and similar

substitutions, together with the assumption that L0 << L will yield the same result for (b). (a) Express the velocity of longitudinal waves on the slinky: ρ

Bv = (1)

For the slinky:

Vm

=ρ

and

VVPB

∆−= (2)

Letting A be the cross-sectional area of the slinky: AL

m=ρ and

ALkP ∆

−=

Chapter 15

1168


ALkB =


mkL

ALmALk

v ==

(b) Express the velocity of transverse waves on the slinky: µ

Fv = (3)

For the slinky:

Lm

=µ

and

( )

LLkLLLkLLLkLkF

<<≈

⎟⎠⎞

⎜⎝⎛ −=−=∆=

0

00

if

1


mkL

LmkLv ==

The Wave Equation 36 •

Picture the Problem The general wave equation is 2

2

22

2 1ty

vxy

∂∂

=∂∂

. To show that each of

the functions satisfies this equation, we’ll need to find their first and second derivatives with respect to x and t and then substitute these derivatives in the wave equation. (a) Find the first two spatial derivatives of ( ) ( ) :3vtxktx,y +=

( )23 vtxkxy

+=∂∂

and

( )vtxkxy

+=∂∂ 62

2

(1)

Find the first two temporal derivatives of ( ) ( ) :3vtxktx,y +=

( )23 vtxkvty

+=∂∂

and

Wave Motion

1169

( )vtxkvty

+=∂∂ 2

2

2

6 (2)

Express the ratio of equation (1) to equation (2):

( )( ) 22

2

2

2

2

166

vvtxkvvtxk

ty

xy

=++

=

∂∂∂∂

confirming that ( ) ( )3, vtxktxy +=

satisfies the general wave equation.

(b) Find the first two spatial derivatives of ( ) ( )vtxikAetx,y −= :

( )vtxikikAexy −=∂∂

and

( )vtxikAekixy −=

∂∂ 22

2

2

or

( )vtxikAekxy −−=

∂∂ 2

2

2

(3)

Find the first two temporal derivatives of ( ) ( )vtxikAetx,y −= :

( )vtxikikvAety −−=∂∂

and

( )vtxikAevkity −=

∂∂ 222

2

2

or

( )vtxikAevkty −−=

∂∂ 22

2

2

(4)


( )

( ) 222

2

2

2

2

2

1vAevk

Aek

ty

xy

vtxik

vtxik

=−−

=

∂∂∂∂

−

−

confirming that ( ) ( )vtxikAetxy −=, satisfies

the general wave equation.

(c) Find the first two spatial derivatives of ( ) ( )vtxktx,y −= ln : vtx

kxy

−=

∂∂

and

( )22

2

2

vtxk

xy

−−=

∂∂

(5)

Chapter 15

1170

Find the first two temporal derivatives of ( ) ( )vtxktx,y −= ln : vtx

vkty

−−=

∂∂

and

( )222

2

2

vtxkv

ty

−−=

∂∂

(6)


( )

( )2

2

22

2

2

2

2

2

2

1v

vtxkvvtx

k

ty

xy

=

−−

−−

=

∂∂∂∂

confirming that ( ) ( )vtxktx,y −= ln

satisfies the general wave equation. *37 •

Picture the Problem The general wave equation is 2

2

22

2 1ty

vxy

∂∂

=∂∂

. To show that

tkxAy ωcossin= satisfies this equation, we’ll need to find the first and second

derivatives of y with respect to x and t and then substitute these derivatives in the wave equation. Find the first two spatial derivatives of tkxAy ωcossin= :

tkxAkxy ωcoscos=∂∂

and

tkxAkxy ωcossin22

2

−=∂∂

(1)

Find the first two temporal derivatives of tkxAy ωcossin= :

tkxAty ωω sinsin−=∂∂

and

tkxAty ωω cossin22

2

−=∂∂

(2)


2

2

2

2

2

2

2

2

2

1

cossincossin

v

ktkxAtkxAk

ty

xy

=

=−−

=

∂∂∂∂

ωωωω

confirming that tkxAy ωcossin=

satisfies the general wave equation.

Wave Motion

1171

Harmonic Waves on a String 38 • Picture the Problem We can find the velocity of the waves from the definition of velocity and their wavelength from .fv=λ

Express the wavelength of the waves: f

v=λ

Using the definition of velocity, find the wave velocity:

m/s12s0.5

m6==

∆∆

=txv

Substitute to obtain: cm20.0

s60m/s12

1 == −λ

39 • Picture the Problem Equation 15-13, ( ) ( ),sin, tkxAtxy ω−= describes a wave

traveling in the positive x direction. For a wave traveling in the negative x direction we have ( ) ( ).sin, tkxAtxy ω+=

(a) Factor k from the argument of the sine function to obtain:

( )

( )vtxkA

tk

xkAx,ty

−=

⎟⎠⎞

⎜⎝⎛ −=

sin

sin ω

(b) Substitute λπ2=k and

fπω 2= to obtain: ( )

⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −=

ftxA

ftxAx,ty

λπ

πλπ

2sin

22sin

(c) Substitute λπ2=k and

Tπω 2= to obtain: ( )

⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −=

tT

xA

tT

Ax,ty

12sin

22sin

λπ

πλπ

Chapter 15

1172

(d) Substitute λπ2=k to obtain: ( )

( )vtxA

txA

txAx,ty

−=

⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −=

λπ

πλω

λπ

ωλπ

2sin

22sin

2sin

(e) Substitute λπ2=k and

fπω 2= to obtain: ( ) ( )

⎟⎠⎞

⎜⎝⎛ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−=

tvxfA

txf

kfA

ftkxAx,ty

π

ππ

π

2sin

22sin

2sin

For waves traveling in the negative x

direction, we simply change the – signs to + signs.

*40 • Picture the Problem We can use f = c/λ to express the frequency of any periodic wave in terms of its wavelength and velocity. (a) Find the frequency of light of wavelength 4×10−7 m:

Hz107.50m104m/s10998.2 14

7

8

×=××

== −λcf

Find the frequency of light of wavelength 7×10−7 m:

Hz1028.4m107m/s10998.2 14

7

8

×=××

== −λcf

Therefore the range of frequencies is:

Hz1050.7Hz1028.4 1414 ×≤≤× f

(b) Use the same relationship to calculate the frequency of these microwaves: Hz101.00

m103m/s10998.2

10

2

8

×=

××

== −λcf

41 • Picture the Problem The average power propagated along the string by a harmonic wave is v,AP 22

21

av µω= where v is the speed of the wave, and µ, ω, and A are the linear

density of the string, the angular frequency of the wave, and the amplitude of the wave,

Wave Motion

1173

respectively. Express and evaluate the power propagated along the string:

vAP 2221

av µω=

The speed of the wave on the string is given by: µ

Fv =

Substitute for v to obtain: µ

µω FAP 2221

av =

Substitute numerical values and evaluate Pav:

( )( )( ) ( ) W87.9kg/m0.05N80m0.05s10kg/m0.054 2212

21

av == −πP

42 • Picture the Problem The average power propagated along the rope by a harmonic wave is v,AP 22

21

av µω= where v is the velocity of the wave, and µ, ω, and A are the linear

density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively. Rewrite the power equation in terms of the frequency of the wave:

vAfvAP 2222221

av 2 µπµω ==

Solve for the frequency:

vP

AvAPf

µπµπav

22av 2

21

2==

The wave velocity is given by:

µFv =

Substitute for v and simplify to obtain: F

PAF

PA

fµπ

µµ

πavav 2

212

21

==


Chapter 15

1174

( )( )

( )Hz171

N60m2kg0.1

W1002m01.02

1=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

πf

43 •• Picture the Problem Equation 15-13, ( ) ( ),sin, tkxAtxy ω−= describes a wave

traveling in the positive x direction. For a wave traveling in the negative x direction, we have ( ) ( )tkxAtxy ω+= sin, . We can determine A, k, and ω by examination of the wave

function. The wavelength, frequency, and period of the wave can, in turn, be determined from k and ω.

(a) direction. negative in the travelingis

wave thepositive, is terms and ebetween thsign theBecausex

tkx ω

Find the speed of the wave:

m/s00.5m8.62s314

1

1

=== −

−

kv ω

(b) The coefficient of x is k and:

λπ2

=k

Solve for and evaluate λ: cm0.10

m8.6222

1 === −

ππλk

The coefficient of t is ω and:

Hz0.502

s3142

1

===−

ππωf

The period of the wave motion is the reciprocal of its frequency:

s0200.0s50

111 === −f

T

(c) Express and evaluate the maximum speed of any string segment:

( )( )m/s0.314

rad/s314m0.001max

=

== ωAv

44 •• Picture the Problem Let the positive x direction be to the right. Then equation 15-13, ( ) ( ),sin tkxAx,ty ω−= describes a wave traveling in the positive x direction. We can

find ω and k from the data included in the problem statement and substitute in the general equation. The maximum speed of a point on the string can be found from ωAv =max and

the maximum acceleration from .2max ωAa =

Wave Motion

1175

(a) Express the general form of the equation of a harmonic wave traveling to the right:

( ) ( )tkxAx,ty ω−= sin

Evaluate ω:

( ) 11 s503s8022 −− === ππω f

Determine k: 11

m9.41m/s12s503 −−

===v

k ω

Substitute to obtain: ( ) ( ) ( ) ( )[ ]txx,ty 11 s503m9.41sinm025.0 −− −=

(b) Express and evaluate the maximum speed of a point on the string:

( )( )m/s12.6

s503m0.025 1max

=

== −ωAv

(c) Express the maximum acceleration of a point on the string:

2max ωAa =


( )( )2

21max

km/s33.6

s503m0.025

=

= −a

45 •• Picture the Problem The average total energy of waves on a string is given by

x,AE ∆=∆ 2221

av µω where µ is the linear density of the string, ω is its angular

frequency, A the amplitude of the wave motion, and, in this problem, ∆x is the length of the string. The average power propagated along the string is .22

21

av vAP µω=

(a) ∆Eav is given by: xAfxAE ∆=∆=∆ 22222

21

av 2 µπµω

Evaluate xAE ∆=∆ 22

21

av µω with ∆x = L = 20 m:

( ) ( ) ( ) J82.6m20m0.012s200m20kg0.062 2212

av =⎟⎟⎠

⎞⎜⎜⎝

⎛=∆ −πE

(b) Express the power transmitted past a given point on the string:

vAP 2221

av µω=

Chapter 15

1176

The speed of the wave is given by:

µFv =

Substitute for v to obtain: µ

µω FAP 2221

av =


( ) ( ) W0.44

m20kg0.06N50m0.012s200

m20kg0.062 2212

av =⎟⎟⎠

⎞⎜⎜⎝

⎛= −πP

*46 •• Picture the Problem The power propagated along the rope by a harmonic wave is

vAP 2221 µω= where v is the velocity of the wave, and µ, ω, and A are the linear density

of the string, the angular frequency of the wave, and the amplitude of the wave, respectively. We can use the wave function ( ) ( )tkxeAy bx ω−= − sin0 to determine the

amplitude of the wave at x = 0 and at point x. (a) Express the power associated with the wave at the origin:

vAP 2221 µω=

Evaluate the amplitude at x = 0:

( ) ( ) 00

00 AeAA ==

Substitute to obtain: ( ) vAP 2

02

210 µω=

(b) Express the amplitude of the wave at x:

( ) ( )bxeAxA −= 0

Substitute to obtain: ( ) ( )bx

bx

veA

veAxP22

02

21

20

221

−

−

=

=

µω

µω

47 ••

Picture the Problem The average power propagated along the rope by a harmonic wave is vAP 22

21

av µω= where v is the velocity of the wave, and µ, ω, and A are the linear

density of the string, the angular frequency of the wave, and the amplitude of the wave, respectively.

Wave Motion

1177

(a) Express the average power transmitted along the wire:

vAfvAP 2222221

av 2 µπµω ==


( )( )( ) ( )

mW0.79

m/s10m100.5

s400kg/m0.01223

212av

=

××

=−

−πP

(b) Because 2

av fP ∝ :

100. offactor aby increase would10 offactor aby Increasing

avPf

Because 2

av AP ∝ :

100. offactor aby increase would10 offactor aby Increasing

avPA

Because vP ∝av and Fv ∝ :

100. offactor aby and 100 offactor aby increase

would10 offactor aby Increasing

av

4

PvF

(c)

easiest. thebe would or increasing source,power

theofity adjustabil on the DependingAf

*48 ••• Picture the Problem We can use the assumption that both the wave function and its first spatial derivative are continuous at x = 0 to establish equations relating A, B, C, k1, and k2. Then, we can solve these simultaneous equations to obtain expressions for B and C in terms of A, v1, and v2. (a) Let y1(x, t) represent the wave function in the region x < 0, and y2(x, t) represent the wave function in the region x > 0. Express the continuity of the two wave functions at x = 0:

( ) ( )tyty ,0,0 21 = and

( )[ ] ( )[ ]( )[ ]tkC

tkBtkAω

ωω−=

++−0sin

0sin0sin

2

11

or ( ) ( )tCtBtA ωωω −=+− sinsinsin

Because the sine function is odd; i.e., ( ) θθ sinsin −=− :

tCtBtA ωωω sinsinsin −=+− and

CBA =− (1)

Chapter 15

1178

Differentiate the wave functions with respect to x to obtain: ( )

( )txkBk

txkAkxy

ω

ω

++

−=∂∂

11

111

cos

cos

and

( )txkCkxy ω−=∂∂

222 cos

Express the continuity of the slopes of the two wave functions at x = 0:

∂y1

∂xx =0

=∂y 2

∂xx =0

and ( )[ ] ( )[ ]

( )[ ]tkCktkBktkAk

ωωω

−=++−

0cos0cos0cos

22

1111

or ( )

( )tCktBktAkω

ωω−=

+−coscoscos

2

11

Because the cosine function is even; i.e., ( ) θθ coscos =− :

tCktBktAk ωωω coscoscos 211 =+ and

CkBkAk 211 =+ (2)

Multiply equation (1) by k1 and add it to equation (2) to obtain:

( )CkkAk 2112 +=

Solve for C: A

kkA

kkkC

1221

1

122

+=

+=

Solve for C/A and substitute ω/v1 for k1 and ω/v2 for k2 to obtain: 2112 1

21

2vvkkA

C+

=+

=


Avv

BA ⎟⎟⎠

⎞⎜⎜⎝

⎛+

=−211

2

Solve for B/A:

21

21

11

vvvv

AB

+−

−=

Wave Motion

1179

(b) We wish to show that B2 + (v1/v2)C2 = A2 Use the results of (a) to obtain the expressions B = −[(1 − α)/(1 + α)] A and C = 2A/(1 + α), where α = v1/v2. Substitute these expressions into B2 + (v1/v2)C2 = A2 and check to see if the resulting equation is an identity:

( )( )

( )

( )( )( )

11

111

1121

11

421

11

41

11

211

12

11

2

2

2

2

2

2

2

2

22

222

22

22

2

12

=

=++

=+++

=+

++−

=+

+−

=⎟⎠⎞

⎜⎝⎛+

+⎟⎠⎞

⎜⎝⎛+−

=⎟⎠⎞

⎜⎝⎛+

+⎟⎠⎞

⎜⎝⎛+−

=+

αα

ααα

αααα

ααα

αα

αα

αα

αα AAA

ACvvB

The equation is an identity: Therefore, 22

2

12 ACvvB =+

Remarks: Our result in (a) can be checked by considering the limit of B/A as v2/v1 → 0. This limit gives B/A = +1, telling us that the transmitted wave has zero amplitude and the incident and reflected waves superpose to give a standing wave with a node at x = 0. Harmonic Sound Waves *49 • Picture the Problem The pressure variation is of the form ( ) ( )vtxkpx,tp −= cos0

where 2π

=k and m/s340=v . We can find λ from k and f from ω and k.

(a) By inspection of the equation: Pa750.00 =p

(b) Because :2

2 πλπ==k

m00.4=λ

(c) Solve kf

kv πω 2

== for f to obtain: ( )Hz0.85

2

m/s3402

2===

π

π

πkvf

Chapter 15

1180

(d) By inspection of the equation: m/s340=v

50 • Picture the Problem The frequency, wavelength, and speed of the sound waves are related by v = fλ. (a) Express and evaluate the wavelength of middle C:

m30.1s262

m/s3401- ===

fvλ

(b) Double the frequency corresponding to middle C; solve for and evaluate λ:

( ) m649.0s2622

m/s3401 === −f

vλ

51 • Picture the Problem The pressure amplitude depends on the density of the medium ρ, the angular frequency of the sound wave µ, the speed of the wave v, and the displacement amplitude s0 according to .00 vsp ρω=

(a) Solve 00 vsp ρω= for s0:

vpsρω

00 =

Substitute numerical values and evaluate s0:

( )( )( )( )( )

m1068.3

m/s340s100kg/m1.292Pa/atm101.01325atm10

5

13

54

0

−

−

−

×=

×=

πs

(b) Use 00 vsp ρω= to find p0:

( )( )( )( ) Pa1027.8m10m/s340s300kg/m29.12 2713

0−−− ×== πp

52 • Picture the Problem The pressure amplitude depends on the density of the medium ρ, the angular frequency of the sound wave µ, the speed of the wave v, and the displacement amplitude s0 according to 00 vsp ρω= .

(a) Solve 00 vsp ρω= for s0:

vpsρω

00 =

Wave Motion

1181

Substitute numerical values and evaluate s0: ( )( )( )

m1010.2

m/s340s1000kg/m1.292Pa29

5

130

−

−

×=

=π

s

(b) Proceed as in (a) with f = 1 kHz:

( )( )( )m1005.1

m/s340s1000kg/m1.292Pa29

5

130

−

−

×=

=π

s

53 • Picture the Problem The pressure or density wave is 90° out of phase with the displacement wave. When the displacement is zero, the pressure and density changes are either a maximum or a minimum. When the displacement is a maximum or minimum, the pressure and density changes are zero. We can use 00 vsp ρω= to find the maximum

value of the displacement at any time and place. (a) If the pressure is a maximum at x1 when t = 0:

zero. is nt displaceme the s

(b) Solve 00 vsp ρω= for s0: v

psρω

00 =

Substitute numerical values and evaluate s0:

( )( )( )( )( )

m68.3

m/s340s1000kg/m1.292Pa/atm101.01325atm10

13

54

0

µ=

×= −

−

πs

*54 • Picture the Problem A human can hear sounds between roughly 20 Hz and 20 kHz; a factor of 1000. An octave represents a change in frequency by a factor of 2. We can evaluate 2N = 1000 to find the number of octaves heard by a person who can hear this range of frequencies. Relate the number of octaves to the difference between 20 kHz and 20 Hz:

10002 =N

Take the logarithm of both sides of the equation to obtain:

310log2log =N or

32log =N

Solve for and evaluate N: 1097.9

2log3

≈==N

Chapter 15

1182

Waves in Three Dimensions: Intensity 55 • Picture the Problem The pressure amplitude depends on the density of the medium ρ, the angular frequency of the sound wave µ, the speed of the wave v, and the displacement amplitude s0 according to .00 vsp ρω= The intensity of the waves is given by

vpvsIρ

ρω20

212

02

21 == and the rate at which energy is delivered (power) is the product of

the intensity and the surface area of the piston. (a) Using ,00 vsp ρω= evaluate p0: ( )( )

( )( )Pa138

m100.1m/s340

s500kg/m1.2923

130

=

××

=−

−πp

(b) Use v

pIρ

20

21= to find the

intensity of the waves:

( )( )( )

2

3

2

W/m7.21

m/s340kg/m1.29Pa138

21

=

=I

(c) Using Pav = IA to find the power required to keep the piston oscillating:

( )( )W217.0

m10W/m21.6 222av

=

= −P

56 • Picture the Problem The intensity of the sound from the spherical source varies inversely with the square of the distance from the source. The power radiated by the source is the product of the intensity of the radiation and the surface area over which it is distributed. (a) Relate the intensity at 10 m to the distance from the source: 2

av

4 rPIπ

=

or

( )2av24

m104W/m10

πP

=−

Letting r′ represent the distance at which the intensity is 10−6 W/m2, express the intensity as in part (a):

2av26

4W/m10

r'Pπ

=−

Wave Motion

1183


2av

2av

26

24

4

m104W/m10W/m10

r'P

P

π

π=−

−

Solve for and evaluate r′: ( )( ) m100m1010 22 ==r'

(b) Solve 2av

4 rPIπ

= for Pav: IrP 2

av 4π=


( ) ( ) W126.0W/m10m104 242av == −πP

*57 • Picture the Problem Because the power radiated by the loudspeaker is the product of the intensity of the sound and the surface area over which it is distributed, we can use this relationship to find the average power, the intensity of the radiation, or the distance to the speaker for a given intensity or average power. (a) Use IrP 2

av 4π= to find the total

acoustic power output of the speaker:

( ) ( )W3.50

W/m10m204 222av

=

= −πP

(b) Relate the intensity of the sound at 20 m to the distance from the speaker:

( )2av22

m204W/m10

πP

=−

Relate the threshold-of-pain intensity to the distance from the speaker:

2av2

4W/m1

rPπ

=

Divide the first of these equations by the second; solve for and evaluate r:

( ) m00.2m2010 22 == −r

(c) Use 2av

4 rPIπ

= to find the

intensity at 30 m:

( )( )

23

2

W/m1045.4

m304W3.50m30

−×=

=π

I

Chapter 15

1184

58 •• Picture the Problem We can use conservation of energy to find the acoustical energy resulting from the dropping of the pin. The power developed can then be found from the given time during which the energy was transformed from mechanical to acoustical form. We can find the range at which the dropped pin can be heard from I = P/4π r2. (a) Assuming that I = P/4π r2, express the distance at which one can hear the dropped pin:

IPrπ4

=

Use conservation of energy to determine the sound energy generated when the pin falls:

( )( )( )( )( )( )

J1091.4m1

m/s81.9kg101.00005.00005.0

7

23

−

−

×=

××=

= mghE

Express the power of the sound pulse:

W104.91s0.1

J104.91

6

7

−

−

×=

×=

∆=

tEP

Substitute numerical values and evaluate r: ( ) m198

W/m104W1091.4

211

6

=×

= −

−

πr

(b) Repeat the last step in (a) with I = 10-8 W/m2: ( ) m25.6

W/m104W1091.4

28

6

=×

= −

−

πr

Intensity Level 59 • Picture the Problem The intensity level of a sound wave β, measured in decibels, is given by ( ) ( )0logdB10 II=β where I0 = 10−12 W/m2 is defined to be the threshold of

hearing. (a) Using its definition, calculate the intensity level of a sound wave whose intensity is 10-10 W/m2:

( )

dB0.2010log10

W/m10W/m10logdB10

2

212

201

==

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

−

β

Wave Motion

1185

(b) Proceed as in (a) with I = 10−2 W/m2: ( )

dB10010log10

W/m10W/m10logdB10

10

212

22

==

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

−

β

60 • Picture the Problem The intensity level of a sound wave β, measured in decibels, is given by ( ) ( )0logdB10 II=β where I0 = 10−12 W/m2 is defined to be the threshold of

hearing. (a) Solve ( ) ( )0logdB10 II=β for I

to obtain:

( )0

dB1010 II β=

Evaluate I for β = 10 dB: ( ) ( )

( ) 211212

00dB10dB10

W/m10W/m1010

1010−− ==

== III

(b) Proceed as in (a) with β = 3 dB: ( ) ( )

( ) 212212

00dB10dB3

W/m102W/m102

210−− ×==

== III

*61 • Picture the Problem The intensity level of a sound wave β, measured in decibels, is given by ( ) ( )0logdB10 II=β where I0 = 10−12 W/m2 is defined to be the threshold of

hearing. Express the sound level of the rock concert: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0

concertconcert logdB10

IIβ (1)

Express the sound level of the dog’s bark: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0

doglogdB10dB50I

I

Solve for the intensity of the dog’s bark:

( )27

21250

5dog

W/m10

W/m101010−

−

=

== II

Express the intensity of the rock concert in terms of the intensity of the dog’s bark:

( )23

274dog

4concert

W/m10

W/m101010−

−

=

== II

Chapter 15

1186

Substitute in equation (1) and evaluate βconcert:

( )

( )dB0.90

10logdB10

W/m10W/m10logdB10

9

212

23

concert

=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

−

β

62 • Picture the Problem The intensity level of a sound wave β, measured in decibels, is given by ( ) ( )0logdB10 II=β where I0 = 10−12 W/m2 is defined to be the threshold of

hearing. Express the intensity level of the louder sound: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0

LL logdB10

IIβ

Express the intensity level of the softer sound: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0

SS logdB10

IIβ

Express the difference between the intensity levels of the two sounds:

( )

( )

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

=−

S

L

0S

0L

0

S

0

L

SL

logdB10

logdB10

loglogdB10

dB30

II

IIII

II

II

ββ

Solve for and evaluate the ratio IL/IS: 3

S

L 10=II

and correct. is )(a

63 • Picture the Problem We can use the definition of the intensity level to express the difference in the intensity levels of two sounds whose intensities differ by a factor of 2. Express the intensity level before the intensity is doubled: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

01 logdB10

IIβ

Express the intensity level with the intensity doubled: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

02

2logdB10IIβ

Wave Motion

1187

Express and evaluate ∆β = β2 − β1:

( ) ( )

( ) ( )

dB0.3dB01.3

2logdB102logdB10

logdB102logdB1000

12

≈=

=⎟⎠⎞

⎜⎝⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−=∆

II

II

II

βββ

*64 • Picture the Problem We can express the intensity levels at both 90 dB and 70 dB in terms of the intensities of the sound at those levels. By subtracting the two expressions, we can solve for the ratio of the intensities at the two levels and then find the fractional change in the intensity that corresponds to a decrease in intensity level from 90 dB to 70 dB. Express the intensity level at 90 dB: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0

90logdB10dB90II

Express the intensity level at 70 dB: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0

70logdB10dB70II

Express ∆β = β90 − β70:

( ) ( )

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟⎠

⎞⎜⎜⎝

⎛=

=∆

70

90

0

70

0

90

logdB10

logdB10logdB10

dB20

II

II

II

β

Solve for I90: 7090 100II =

Express the fractional change in the intensity from 90 dB to 70 dB:

%99100

100

70

7070

90

7090 =−

=−

III

III

65 •• Picture the Problem The intensity at a distance r from a spherical source varies with distance from the source according to .4 2

av rPI π= We can use this relationship to

relate the intensities corresponding to an 80-dB intensity level (I80) and the intensity corresponding to a 60-dB intensity level (I60) to their distances from the source. We can relate the intensities to the intensity levels through ( ) ( )0logdB10 II=β .

Chapter 15

1188

(a) Express the intensity of the sound where the intensity level is 80 dB:

210

av10 4 r

PIπ

=

Express the intensity of the sound where the intensity level is 60 dB:

2av

60 4 rPIπ

=


( )2

2

2av

2av

60

80

m1004

m104 r

rP

P

II

==

π

π

Solve for r: ( )

60

80m10IIr =

Find the intensity of the 80-dB sound level radiation: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0

80logdB10dB80II

and 24

08

80 W/m1010 −== II

Find the intensity of the 60-dB sound level radiation: ( ) ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

0

60logdB10dB60II

and 26

06

60 W/m1010 −== II

Substitute and evaluate r:

( ) m100W/m10W/m10m10 26

24

== −

−

r

(b) Using the intensity corresponding to an intensity level of 80 dB, express and evaluate the power radiated by this source:

( ) ( )[ ]W126.0

m104πW/m10 224

80

=

=

=−

AIP

66 •• Picture the Problem Let I1 and I2 be the intensities of the sound at distances r1 and r2. We can relate these intensities to the intensity levels through

( ) ( )0logdB10 II=β and to the distances through .4 2av rPI π=

Wave Motion

1189

Using ( ) ( )0logdB10 II=β ,

express the ratio β1/β2:

( ) ( )( ) ( )

01

02

01

02

1

2

logloglogloglogdB10logdB10

IIII

IIII

−−

=

=ββ

Express the ratio of the intensities at distances r1 and r2 from the source and solve for I2:

22

21

1

2

rr

II

= and 122

21

2 IrrI =


( )

( ) ( )( )

( )1

211

01

2101

01

0211

01

0122

21

1

2

log20log10

log20log10

log10log10log10log20log10

loglog

log10log10

ββ

ββ

rrII

rrII

IIIrrI

II

IIrr

+=

+=

−−+

=−

−=

67 •• Picture the Problem We can use ( ) ( )0logdB10 II=β where I0 = 10−12 W/m2 is

defined to be the threshold of hearing, to find the intensity level at 20 m. Because the power radiated by the loudspeaker is the product of the intensity of the sound and the surface area over which it is distributed, we can use this relationship to find either the average power, the intensity of the radiation, or the distance to the speaker for a given intensity or average power. (a) Relate the intensity level to the intensity at 20 m:

( )

( )

( ) ( ) dB10010logdB10

W/m10W/m10logdB10

logdB10

10

212

22

0

==

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

IIβ

(b) Use IrP 2

av 4π= to find the total

acoustic power output of the speaker:

( ) ( )W3.50

W/m10m204 222av

=

= −πP

(c) Relate the intensity of the sound at 20 m to the distance from the speaker:

( )2av22

m204W/m10

πP

=−

Chapter 15

1190

Relate the threshold-of-pain intensity to the distance from the speaker:

2av2

4W/m1

rPπ

=

Divide the first of these equations by the second; solve for and evaluate r:

( ) m00.2m2010 22 == −r

(d) Use 2av

4 rPIπ

= to find the

intensity at 30 m:

( )( )

23

2

W/m1045.4m304W3.50m30

−×=

=π

I

Find the intensity level at 30 m: ( ) ( )

( )dB5.96

104.45logdB10W/m10

W/m104.45logdB10m30

9

221

23

=

×=

×= −

−

β

68 •• Picture the Problem Let I′ and I represent the sound in consecutive years. Then, we can use ( ) ( )0logdB10 II=β to express the annual increase in intensity levels.

(a) Express the annual change in intensity level: ( ) ( )

( )II'

II

II'

'

logdB10

logdB10logdB10

dB1

00

=

−=

=−=∆ βββ

Solve for I′/I: 26.110 1.0 ==

II'

and the annual increase in intensity is %.26 This is not a plausible annual

increase because, if it were true, the intensity level would increase by a factor of 10 in ten years.

(b) Because doubling of the intensity corresponds to ∆β = 3 dB and the intensity is increasing 1 dB annually:

years. 3in double willlevelintensity The

Wave Motion

1191

69 •• Picture the Problem We can find the intensities of the three sources from their intensity levels and, because their intensities are additive, find the intensity level when all three sources are acting. (a) Express the sound intensity level when the three sources act at the same time:

( )

( )0

807370

0

sources3sources3

logdB10

logdB10

IIII

II

++=

=β

Find the intensities of each of the three sources:

( ) 07

700

70 10logdB10dB70 IIII

=⇒=

( ) 03.7

730


=⇒=

and

( ) 08

800


=⇒=

Substitute and evaluate β3 sources:

( ) ( ) ( )

dB1.81

101010logdB10101010logdB10 83.77

0

08

03.7

07

sources3

=

++=++

=I

IIIβ

(b) Find the intensity level with the two least intense sources eliminated:

( ) ( ) ( )

dB0.80

10logdB1010logdB10 8

0

08

80

=

==I

Iβ

tly.significan levelintensity thereducenot does sources intenseleast two thegEliminatin

*70 •• Picture the Problem Let P be the power radiated by the source of sound, and r be the initial distance from the source to the receiver. We can use the definition of intensity to find the ratio of the intensities before and after the distance is doubled and then use the definition of the decibel level to find the change in its level. Relate the change in decibel level to the change in the intensity level:

'log10

II

=∆β

Chapter 15

1192

Using its definition, express the intensity of the sound from the source as a function of P and r:

24 rPIπ

=

Express the intensity when the distance is doubled:

( ) 22 1624 rP

rPI'

ππ==

Evaluate the ratio of I to I′:

4

16

4

2

2

==

rPr

P

I'I

π

π

Substitute to obtain: dB02.64log10 ==∆β

and correct. is )(c

71 ••• Picture the Problem The sound level can be found from the intensity of the sound due to the talking people. When 38 people are talking, the intensities add. Express the sound level when all 38 people are talking:

( )

( ) ( )

( )dB8.87

dB7238logdB10

logdB1038logdB10

38logdB10

0

1

0

138

=

+=

+=

=

II

IIβ

An equivalent but longer solution:

Express the sound level when all 38 people are talking:

( )0

138

38logdB10I

I=β

Express the sound level when only one person is talking:

( )0

11 logdB10dB72

II

==β

Solve for and evaluate I1: ( )

25

2122.70

2.71

W/m101.58

W/m101010−

−

×=

== II

Express the intensity when all 38 people are talking:

138 38II =

Wave Motion

1193

The decibel level is: ( ) ( )

dB8.87

W/m10W/m101.5838logdB10 212

25

38

=

×= −

−

β

*72 ••• Picture the Problem Let η represent the efficiency with which mechanical energy is converted to sound energy. Because we’re given information regarding the rate at which mechanical energy is delivered to the string and the rate at which sound energy arrives at the location of the listener, we’ll take the efficiency to be the ratio of the sound power delivered to the listener divided by the power delivered to the string. We can calculate the power input directly from the given data. We’ll calculate the intensity of the sound at 35 m from its intensity level at that distance and use this result to find the power output. Express the efficiency of the conversion of mechanical energy to sound energy:

in

out

PP

=η

Find the power delivered by the bow to the string:

( )( ) W0.3m/s0.5N0.6in === FvP

Using ( ) ( )0logdB10 II=β , find

the intensity of the sound at 35 m:

( )0

m35logdB10dB60I

I=

and 26

06

m35 W/m1010 −== II

Find the power of the sound emitted:

( )( )W0.0154

m35W/m104π 226out

=== −IAP

Substitute numerical values and evaluate η:

%13.5W0.3

W0.0154==η

73 ••• Picture the Problem Because the sound intensities are additive, we’ll find the intensity due to one student by subtracting the background intensity from the intensity due to the students and dividing by 100. Then, we’ll use this result to calculate the intensity level due to 50 students. Express the intensity level due to 50 students:

( )0

150

50logdB10I

I=β

Chapter 15

1194

Find the sound intensity when 100 students are writing the exam:

( )0

100logdB10dB60I

I=

and 26

06

100 W/m1010 −== II

Find the sound intensity due to the background noise:

( )0

backgroundlogdB10dB40I

I=

and 28

04

background W/m1010 −== II

Express the sound intensity due to the 100 students: 26

2826background100

W/m10

W/m10W/m10−

−−

≈

−=− II

Find the sound intensity due to 1 student:

28background100 W/m10100

−=− II

Substitute numerical values and evaluate the intensity level due to 50 students:

( ) ( )

dB0.57

W/m10W/m1050logdB10 212

28

50

=

= −

−

β

The Doppler Effect 74 •

Picture the Problem We can use equation 15-32 (sfuv ±

=λ ) to find the wavelength of

the sound between the source and the listener and 15-35a ( ss

rr f

uvuvf

±±

= ) to find the

frequency heard by the listener. (a) Because the source is approaching the listener, use the minus sign in the numerator of Equation 15-32 to find the wavelength of the sound between the source and the listener:

m30.1s200

m/s80m/s3401 =

−= −λ

Wave Motion

1195

(b) Because the listener is at rest and the source is approaching, ur = 0 and the denominator of Equation 15-35a is the difference between the two speeds:

( )

Hz262

s200m/s80m/s340

m/s340 1

ss

r

=

−=

−=

−

fuv

vf

75 • Picture the Problem In the reference frame described, the speed of sound from the source to the listener is reduced by the speed of the wind. We can find the wavelength of the sound in the region between the source and the listener from v = fλ. Because the sound waves in the region between the source and the listener will be compressed by the motion of the listener, the frequency of the sound heard by the listener will be higher than

the frequency emitted by the source and can be calculated using ss

rr f

uvuvf

±±

= .

(a) The speed of sound in the reference frame of the source is:

m/s260m/s80m/s403 =−=v

(b) Noting that the frequency is unchanged, express the wavelength of the sound:

m30.1s200m/s260

1 === −fvλ

(c) Apply Equation 15-35a to obtain:

( )

Hz262

s200m/s260

m/s80m/s260 1

sr

r

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

⎟⎠⎞

⎜⎝⎛ +

=

−

fvuvf

76 • Picture the Problem We can use ( ) sfuv ±=λ to find the wavelength of the sound in

the region between the source and the listener and ss

rr f

uvuvf

±±

= to find the frequency

heard by the listener. Because the sound waves in the region between the source and the listener will be spread out by the motion of the listener, the frequency of the sound heard by the listener will be lower than the frequency emitted by the source.

Chapter 15

1196

(a) Because the source is moving away from the listener, use the positive sign in the numerator of Equation 15-32 to find the wavelength of the sound between the source and the listener:

m10.2

s200m/s80m/s340

1

=

+= −λ

(b) Because the listener is at rest and the source is receding, ur = 0 and the denominator of Equation 15-35a is the sum of the two speeds: ( )

Hz162

s200m/s80m/s340

m/s340 1

ss

r

=

+=

+=

−

fuv

vf

77 • Picture the Problem We can find the wavelength of the sound in the region between the source and the listener from v = fλ. Because the sound waves in the region between the source and the listener will be compressed by the motion of the listener, the frequency of the sound heard by the listener will be higher than the frequency emitted by the source

and can be calculated using ss

rr f

uvuvf

±±

= .

(a) Because the wavelength is unaffected by the motion of the observer:

m1.70s200m/s340

1 === −fvλ

(b) Apply Equation 15-35a to obtain:

( )

Hz247

s200m/s403

m/s80m/s403 1

0r

r

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

⎟⎠⎞

⎜⎝⎛ +

=

−

fvuvf

78 • Picture the Problem In this reference frame, the speed of sound will be increased by the speed of the listener. We can find the wavelength of the sound in the region between the source and the listener from v = fλ. Because the sound waves in the region between the source and the listener will be compressed by the motion of the listener, the frequency of the sound heard by the listener will be higher than the frequency emitted by the source and can be calculated using v′ = frλ′.

Wave Motion

1197

(a) Moving at 80 m/s in still air: m/s.80 of velocity wind

a sexperienceobserver The

(b) Use the standard Galilean transformation to obtain: m/s420

m/s80m/s340r

=

+=+= uvv'

(c) Because the distance between the wave crests is unchanged:

m70.1s200m/s340

1 === −fv''λ

(d) Using the speed of sound in this reference frame, express and evaluate the frequency heard by the listener:

Hz247m1.70

m/s420r ===

'v'fλ

79 • Picture the Problem Because the listener is moving away from the source, we know that the frequency he/she will hear will be less than the frequency emitted by the source. We

can use ss

rr f

uvuvf

±±

= , with us = 0 and the minus sign in the numerator, to determine its

value. Relate the frequency heard by the listener to that of the source:

( )

Hz153

s200m/s340

m/s80m/s340 1

sr

r

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

⎟⎠⎞

⎜⎝⎛ −

=

−

fvuvf

80 • Picture the Problem The diagram shows the position of the supersonic plane at time t after it was directly over a person located at point P 5000 m below it. Let u represent the speed of the plane and v the speed of sound. We can use trigonometry to determine the angle of the shock wave as well as the location of the jet x when the person on the ground hears the shock wave.

Chapter 15

1198

(a) Referring to the diagram, express θ in terms of v, u, and t:

°=

⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛= −−−

6.23

5.21sin1sinsin 111

vuutvtθ

(b) Using the diagram, relate the angle θ to the altitude h of the plane and the distance x and solve for x:

xh

=θtan ⇒ θtan

hx =


km4.11tan23.6

m5000=

°=x

81 • Picture the Problem If both us and ur are much smaller than the speed of sound v, then

the shift in frequency is given approximately by ,r v

uff

±=∆

where u = us ± ur is the

relative speed of the source and receiver. For purposes of this problem, we’ll assume that you are an Olympics qualifier and can run at a top speed of approximately 10 m/s. Express the frequency, fr, you hear in terms of the frequency of the source fs and your running speed u:

sr fv

uvf ⎟⎠⎞

⎜⎝⎛ +

=

Assuming that you can run 10 m/s , substitute numerical values and evaluate fr:

( )

Hz1029

Hz1000m/s340

m/s10m/s340r

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=f

Using the positive sign (you are approaching the source), express

and evaluate the ratio rff∆

:

%94.2m/s340

m/s01

r

==∆ff

speed. runningyour estimate pitch to of senseyour use to be it would frequency,in change a ofn recognitio

for criterion 3% than theless isfrequency in change fractional thisBecauseimpossible

82 •• Picture the Problem Because the car is moving away from the radar device, the frequency fr it receives will be less than the frequency emitted by the device. The

Wave Motion

1199

microwaves reflected from the car, moving away from a stationary detector, will be of a still lower frequency fr ′. We can use the Doppler shift equations to derive an expression for the speed of the car in terms of difference of these frequencies. Express the frequency fr received by the moving car in terms of fs, u, and v:

sr fc

ucf ⎟⎠⎞

⎜⎝⎛ −

= (1)

Express the frequency fr ′ received by the stationary source in terms of fr, u, and v:

rr fc

uc'f ⎟⎠⎞

⎜⎝⎛ −

= (2)

Substitute equation (1) in equation (2) to eliminate fr:

ss

2

r21 fcuf

cuc'f ⎟

⎠⎞

⎜⎝⎛ −≈⎟

⎠⎞

⎜⎝⎛ −

=

provided u << c

Express the frequency difference detected at the source:

s

ssrs

2

21'

fcu

fcuffff

=

⎟⎠⎞

⎜⎝⎛ −−=−=∆

Solve for u: f

fcu ∆=

s2

Substitute numerical values and evaluate u: ( ) ( )

km/h1.79

hs3600

m10km1

sm0.22

Hz293GHz22

m/s103

3

8

=

××=

×=u

*83 •• Picture the Problem Because the radial component u of the velocity of the raindrops is small compared to the speed v = c of the radar pulse, we can approximate the fractional change in the frequency of the reflected radar pulse to find the speed of the winds carrying the raindrops in the storm system. Express the shift in frequency when the speed of the source (the storm system) u is much smaller than the wave speed v = c:

cu

ff≈

∆

s

Chapter 15

1200

Solve for u: sf

fcu ∆=


( )

mi/h493

m/s0.2778mi/h0.6215m/s156

MHz625Hz325m/s10998.2 8

=

×=

×=u

84 •• Picture the Problem Let the depth of the submarine be represented by D and its vertical speed by u. The submarine acts as both a receiver and source. We can apply the definition of average speed to determine the depth of the submarine and use the Doppler shift equations to derive an expression for the vertical speed of the submarine in terms of the frequency difference. (a) Using the definition of average speed, relate the depth of the submarine to the time delay between the transmitted and reflected pulses:

tvD ∆=2 ⇒ tvD ∆= 21

Substitute numerical values and evaluate D:

( )( ) m61.6ms80km/s1.5421 ==D

(b) Express the frequency fr received by the submarine in terms of f0, u, and c:

sr fc

ucf ⎟⎠⎞

⎜⎝⎛ ±

= (1)

Express the frequency fr′ received by the destroyer in terms of fr, u, and c:

rr fc

uc'f ⎟⎠⎞

⎜⎝⎛ ±

= (2)


ss

2

r21' fcuf

cucf ⎟

⎠⎞

⎜⎝⎛ ±≈⎟

⎠⎞

⎜⎝⎛ ±

=

provided u << c.

Express the frequency difference detected by the destroyer:

s

ssrs

2

21

fcu

fcuf'fff

=

⎟⎠⎞

⎜⎝⎛ ±−=−=∆

Wave Motion

1201

Solve for u: ff

cu ∆=s2

Substitute numerical values and evaluate u: ( ) ( )

m/s809.0

MHz0420.0MHz042km/s54.1

=

=u

where the positive speed indicates that the velocity of the submarine is downward.

85 •• Picture the Problem Because the car is moving away from the radar unit, the frequency fr it receives will be less than the frequency emitted by the unit. The radar waves reflected from the car, moving away from a stationary detector, will be of a still lower frequency fr′. Let u represent the relative speed of the police car and the receding car (140 km/h) and use the Doppler shift equations to derive an expression for the difference between fs, the transmitted signal, and fr′. Express the frequency fr received by the moving car in terms of fs, u, and c:

sr fc

ucf ⎟⎠⎞

⎜⎝⎛ −

= (1)

Express the frequency fr′ received by the stationary source in terms of fr, u, and c:

rr fc

ucf' ⎟⎠⎞

⎜⎝⎛ −

= (2)


ss

2

r211 fcuf

cu'f ⎟

⎠⎞

⎜⎝⎛ −≈⎟

⎠⎞

⎜⎝⎛ −=

provided u << c


s

ssrs

2

21

fcu

fcuff'ff

=

⎟⎠⎞

⎜⎝⎛ −−=−=∆

Substitute numerical values and evaluate ∆f: ( )

kHz78.7

Hz103m/s103

s3600h1

hkm1402

108

=

×

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×

⎟⎟⎠

⎞⎜⎜⎝

⎛×

=∆f

Chapter 15

1202

86 •• Picture the Problem Because the car is moving away from the radar unit, the frequency fr it receives will be less than the frequency emitted by the unit. The radar waves reflected from the car, moving away from a stationary detector, will be of a still lower frequency fr′. Let u represent the relative speed of the police car and the receding car (80 km/h) and use the Doppler shift equations to derive an expression for the difference between fs, the transmitted signal, and fr′. Express the frequency fr received by the moving car in terms of fs, u, and c:

sr fc

ucf ⎟⎠⎞

⎜⎝⎛ −

= (1)

Express the frequency fr′ received by the stationary source in terms of fr, u, and c:

rr fc

uc'f ⎟⎠⎞

⎜⎝⎛ −

= (2)


ss

2

r211 fcuf

cu'f ⎟

⎠⎞

⎜⎝⎛ −≈⎟

⎠⎞

⎜⎝⎛ −=

provided u << c.


s

0srs

2

s21

fcu

fcuf'fff

=

⎟⎠⎞

⎜⎝⎛ −−=−=∆

Substitute numerical values and evaluate ∆f: ( )

kHz44.4

Hz103m/s103

s3600h1

hkm802

108

=

×

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×

⎟⎟⎠

⎞⎜⎜⎝

⎛×

=∆f

Wave Motion

1203

87 •• Picture the Problem The diagram shows the position of the supersonic plane flying due west at time t after it was directly over point P 12 km below it. Let x be measured from point P, u represent the speed of the plane, and v be the speed of sound. We can use trigonometry to determine the angle of the shock wave as well as the location of the jet x when the person on the ground hears the shock wave.

Using the diagram to relate the distance x to the shock-wave angle θ and the elevation of the plane:

xh

=θtan and θtan

hx =

Referring to the diagram, express θ in terms of v, u, and t and determine its value:

°=⎟⎠⎞

⎜⎝⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛=

−

−−

7.386.1

1sin

1sinsin

1

11

vuutvtθ


. of west km0.15tan38.7

km12 Px =°

=

88 •• Picture the Problem The change in frequency of source as it oscillates on the air track is

3 Hz. We can use vu

ff

±≈∆

s

to find the maximum speed of the vibrating mass-spring

system in terms of this change in frequency and then use this speed to find the energy of the system. Knowing the energy of the system, we can find the amplitude of its motion. We can calculate the period of the motion from knowledge of the mass of the radio and the stiffness constant of the spring. (a) Express the energy of the vibrating mass-spring system in terms of its maximum speed:

2max2

1 muE = (1)

Relate the change in the frequency heard by the listener to the maximum speed of the oscillator:

vu

ff

±=∆

s

where u = us ± ur is the relative speed of the source and receiver and v is the speed of

Chapter 15

1204

sound.

Solve for and evaluate u = umax: ( )

m/s1.275

m/s340Hz800

Hz3

0max

=

=∆

= vffu

Substitute numerical values in equation (1) and evaluate E:

( )( ) mJ81.3m/s1.275kg0.1 2221 ==E

(b) Relate the energy of the oscillator to the amplitude of its motion:

221 kAE =

Solve for A to obtain:

kEA 2

=


( ) cm2.85N/m200

mJ81.32==A

89 •• Picture the Problem The received and transmitted frequencies are related through

ss

rr f

uvuvf

±±

= , where the variables have the meanings given in the problem statement.

Because the source and receiver are moving in the same direction, we use the minus signs in both the numerator and denominator. (a) Relate the received frequency f r to the frequency f0 of the source: ( )( ) 0

1sr

0s

rr

11

11

fvuvu

fvuvuf

−−−=

−−

=

(b) Apply the binomial expansion to ( ) 1

s1 −− vu :

( ) vuvu s1

s 11 +≈− −


( )( )( )( )[ ]

⎟⎠⎞

⎜⎝⎛ −+≈

−−+=+−=

vuu

fvuvuvuvufvuvuf

rs

0srrs

0srr

1

111

because both us and ur are small compared to v.

Wave Motion

1205

Because urel = us − ur 0

relr 1 f

vu'f ⎟

⎠⎞

⎜⎝⎛ +≈

90 •• Picture the Problem Because the students are walking away from each other, the frequency f ′ each receives will be less than the frequency fs = 440 Hz emitted by their tuning forks. Let u represent the speed of each student and use the Doppler shift equations to derive an expression for the difference between fs, the frequency of the tuning fork each carries, and the frequency heard from the other’s tuning fork. Because this equation will contain u, we’ll be able to solve for and evaluate each student’s walking speed. Using equation 15-35a, express the frequency fr received by either student, when they are walking away from each other, in terms of fs, u, and v:

( )( ) s1

sr

11

11

fvuvu

fvuvuf

−+−=

+−

=

Expand ( ) 11 −+ vu binomially: ( ) vuvu −≈+ − 11 1 provided u << v.


( )( )( ) s

s22

s2

r

2121

1

fvufvuvu

fvuf

−≈+−=

−=

for u << v.

Express the frequency difference heard by each student:

sssr0221 fvuf

vuffff =⎟⎠⎞

⎜⎝⎛ −−=−=∆

Solve for u: v

ffu

s2∆

=

Substitute numerical values and evaluate u: ( ) ( ) m/s0.773m/s340

Hz4402Hz2

==u

91 •• Picture the Problem The student serves as a source moving toward the wall, and a moving receiver for the echo. The received and transmitted frequencies are related

through ss

rr f

uvuvf

±±

= . We’ll express the frequency received by the wall and the

frequency it transmits back to the moving student in order to express the difference in the

Chapter 15

1206

frequency the student hears from the wall and frequency she hears directly from her tuning fork. Because this expression will contain the student’s walking speed, we’ll be able to solve for and evaluate this speed. Relate the frequency fr received by the wall to the frequency of the student’s tuning fork fs: s

ssrs

2

21

fcu

fcuf'fff

=

⎟⎠⎞

⎜⎝⎛ −−=−=∆

Because the source is moving toward a stationary receiver:

( ) s1

sss

r 11

1 fvufvu

f −−=−

=

Apply the binomial expansion to ( ) 1

s1 −− vu :

( ) vuvu s1

s 11 +≈− −

because us << v.


( ) 0sr 1 fvuf +=

Relate the frequency fr′ reflected from the wall and received by the student to the frequency fr reflected from the wall:

rs

rs

rr 1

111 f

vuf

vuvu'f

−=

±±

=

because the source (the wall) is at rest and the receiver is approaching.

Substitute for fr to obtain: ( )

( )( ) s1

ss

sss

r

11

11

1

fvuvu

fvuvu

'f

−−+=

+−

=

Expand ( ) 1

s1 −− vu binomially: ( ) vuvu s1

s 11 +≈− −

because us << v.

( )( )( ) ss

sssr

2111

fvufvuvu'f

+≈++=

because us << v

Express the difference between the frequency the student receives from the wall and the frequency of her tuning fork:

( )

ss

sss

sr

221

fvu

ffvuf'ff

=

−+=−=∆

Wave Motion

1207

Solve for us: vffu

ss 2

∆=

Substitute numerical values and evaluate us: ( ) ( ) m/s1.33m/s340

Hz5122Hz4

s ==u

*92 •• Picture the Problem The frequency heard by the stationary observer will vary with time as the speaker rotates on its support arm. We can use a Doppler equation to express the frequency heard by the observer as a function of the velocity of the source and find the velocity of the source from the expression for the tangential velocity of an object moving in a circular path. Express the frequency fr heard by a stationary observer:

( ) s1

sss

r 11

1 fvufvu

f −−=−

=

Expand ( ) 1

s1 −− vu to obtain: ( ) vuvu s1

s 11 +≈− −

because us/v << 1

Substitute in the expression for fr:

( ) ssr 1 fvuf += (1)

Express the speed of the source as a function of time:

( )( ) ( )[ ]( ) ( )[ ]t

ttru

rad/s4sinm/s2.3rad/s4sinrad/s4m0.8

sins

=== ωω

Substitute in equation (1) to obtain: ( )[ ] sr rad/s4sinm/s3.21 ft

vf ⎟

⎠⎞

⎜⎝⎛ +=

Substitute for v and simplify: ( )[ ] ( )

( ) ( )[ ]t

tf

rad/s4sinHz9.41Hz1000

s1000rad/s4sinm/s340m/s3.21 1

r

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+= −

93 •• Picture the Problem The simplest way to approach this problem is to transform to a reference frame in which the balloon is at rest. In that reference frame, the speed of sound is v = 340 m/s, and ur = 36 km/h = 10 m/s. Then, we can use the equations for a moving receiver and a moving source to find the frequencies heard at the window and on the balloon.

Chapter 15

1208

(a) Express the observed frequency in terms of the frequency of the source:

sr

r 1 fvuf ⎟⎠⎞

⎜⎝⎛ +=

Substitute numerical values and evaluate fr:

( ) Hz824Hz800m/s340

m/s101r =⎟⎟⎠

⎞⎜⎜⎝

⎛+=f

(b) Treating the tall building as a moving source, express the frequency of the reflected sound heard by a person riding in the balloon:

rs

r

1

1 f

vu'f⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−=

Substitute numerical values and evaluate fr′: ( ) Hz849Hz824

m/s340m/s101

1r =

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−='f

94 •• Picture the Problem We can relate the frequencies fr and fr′ heard by the stationary observer behind the car to the speed of the car u and the frequency of the car’s horn fs. Dividing these equations will eliminate the frequency of the car’s horn and allow us to solve for the speed of the car. We can then substitute to find the frequency of the car’s horn. We can find the frequency heard by the driver as a moving receiver by relating this frequency to the frequency reflected from the wall. (a) Relate the frequency heard by the observer directly from the car’s horn to the speed of the car:

sr 11 f

vuf

+= (1)

Relate the frequency reflected from the wall to the speed of the car: sr 1

1 fvu

'f−

= (2)

Divide equation (2) by equation (1) to obtain: vu

vuf'f

−+

=11

r

r

Solve for u: v

'f'ff'furr

rr

+−

=

Wave Motion

1209


( )

km/h8.89

hs3600

m10km1

sm95.24

m/s340Hz745Hz863Hz745Hz863

3

=

××=

+−

=u

(b) Solve equation (1) for fs: ( ) rs 1 fvuf +=

Substitute numerical values and evaluate fs:

( )

Hz800

Hz745m/s340m/s24.951s

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=f

(c) The driver is a moving receiver and so we can relate the frequency heard by the driver to the frequency reflected by the wall (the frequency heard by the stationary observer):

'fvuf rdriver 1 ⎟⎠⎞

⎜⎝⎛ +=

Substitute numerical values and evaluate fdriver:

( )

Hz926

H863m/s340m/s24.951driver

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+= zf

95 •• Picture the Problem Let t = 0 when the driver sounds her horn and let the distance to the cliff at that instant be d. The received and transmitted frequencies are related

through ss

rr f

uvuvf

±±

= . Solving this equation for fs will allow us to determine the

frequency of the car horn. We can use the total distance the sound travels (car-to-cliff plus cliff back to car … now closer to the cliff) to determine the distance to the cliff when the horn was briefly sounded. Relate the frequency heard by the driver to her speed and to the frequency of her horn:

ss

rr 1

1 fvuvuf

−+

=

Solve for fs: r

r

ss 1

1 fvuvuf

+−

=

Chapter 15

1210


( ) ( ) Hz713Hz840

m/s340m/s78.271

m/s340m/s78.271

Hz840

m/s340s3600

h1h

km1001

m/s340s3600

h1h

km1001

0 =+

−=

×+

×−

=f

Relate the distance d to the cliff at t = 0 to the distance she travels in time ∆t = 1 s, her speed u, and the speed of sound v:

( ) tvtudd ∆=∆−+

Solve for d:

( ) tvud ∆+= 21


( )( )m184

s1m/s340m/s27.7821

=

+=d

96 •• Picture the Problem You’ll year the sonic boom when the surface of its cone reaches your plane. In the diagram the Concorde is at C and your plane is at P. The distance h = 3 km. The distance between the planes when you hear the sonic boom is d. We can use trigonometry to determine the angle of the shock wave as well as the separation of the planes when you hear the sonic boom.

Using the Pythagorean theorem, relate the separation of the planes d, to the distance h and the angle θ :

θ2222 cosdhd +=

Solve for d to obtain: θ2cos1

1−

= hd

Express θ in terms of v, u, and t:

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛= −−

vuutvt 1sinsin 11θ

Wave Motion

1211


°=⎟⎠⎞

⎜⎝⎛= − 7.38

6.11sin 1θ

Substitute numerical values and evaluate h: ( ) km80.4

7.38cos11km3 2 =

°−=d

*97 •• Picture the Problem The sun and Jupiter orbit about their effective mass located at their common center of mass. We can apply Newton’s 2nd law to the sun to obtain an expression for its orbital speed about the sun-Jupiter center of mass and then use this speed in the Doppler shift equation to estimate the maximum and minimum wavelengths resulting from the Jupiter-induced motion of the sun. Letting v be the orbital speed of the sun about the center of mass of the sun-Jupiter system, express the Doppler shift of the light due to this motion when the sun is approaching the earth:

cvcvc

cvcvf

'cf'

/1/1

/1/1

−+

=−+

==λλ

Solve for λ′:

( )( )( ) ( ) 2121

1

11

11

/1/1

−

−

+−=

+−=

+−

=

cvcv

cvcv

cvcv'

λ

λ

λλ

Because v << c, we can expand ( ) 211 cv− and ( ) 211 −+ cv

binomially to obtain:

( )c

vcv2

11 21 −≈−

and

( )c

vcv2

11 21 −≈+ −


cv

cv

cvcv

−≈⎟⎠⎞

⎜⎝⎛ −=

+− 1

21

/1/1 2

When the sun is receding from the earth:

cv

cv

cvcv

+≈⎟⎠⎞

⎜⎝⎛ +=

−+ 1

21

/1/1 2

Hence the motion of the sun will give an observed Doppler shift of:

⎟⎠⎞

⎜⎝⎛ ±≈

cv' 1λλ (1)

Chapter 15

1212

Apply Newton’s 2nd law to the sun:

cm

2

S2cm

effS

rvM

rMGM

=


cm

eff

rGMv =

Measured from the center of the sun, the distance to the center of mass of the sun-Jupiter system is:

( )Js

JJ-s

Js

JJ-sScm

0MM

MrMM

MrMr+

=++

=

The effective mass is related to the masses of the sun and Jupiter according to:

Jseff

111MMM

+=

or

Js

Jseff MM

MMM+

=

Substitute for Meff and rcm to obtain:

J-s

s

Js

JJ-s

Js

Js

rGM

MMMr

MMMMG

v =

+

+=

Using m1078.7 11

J-s ×=r as the mean orbital radius of Jupiter, substitute numerical

values and evaluate v:

( )( ) m/s10306.1m1078.7

kg1099.1kg/mN10673.6 411

302211

×=×

×⋅×=

−

v


( )( )5

8

4

1036.41nm500

m/s10998.2m/s10306.11nm500'

−×±=

⎟⎟⎠

⎞⎜⎜⎝

⎛××

±≈λ

The maximum and minimum wavelengths are:

( )( )5max 1036.41nm500 −×+=λ

and ( )( )5

min 1036.41nm500 −×−=λ

98 ••• Picture the Problem Choose a coordinate system in which downward is the positive y direction. Let d represent the distance the tuning fork has fallen when the student hears a

Wave Motion

1213

frequency of 400 Hz, t1 the time for the source to fall that distance, and t2 the time for the sound to travel back to the student. We can a constant-acceleration equation to express d in terms of the time that elapses between the dropping of the tuning fork and the return of the sound to the student. A Doppler-effect equation will allow us to solve for the speed of the tuning fork when the student hears a frequency of 400 Hz; we can use constant-acceleration equations to find the fall time for the fork and the return time for the sound from the tuning fork. Using a constant-acceleration equation, relate the distance the source has fallen to the elapsed time:

221

0 attvd +=

or, because v0 = 0 and a = g, 2

21 gtd = (1)

where t = t1 + t2.

Relate the frequency fr heard by the student to the speed of the falling tuning fork:

ss

r 11 f

vuf

+=

Solve for us: v

ffu ⎟⎟

⎠

⎞⎜⎜⎝

⎛−= 1

r

ss

Substitute numerical values and evaluate us:

( ) m/s34.0m/s3401Hz400Hz440

s =⎟⎟⎠

⎞⎜⎜⎝

⎛−=u

Letting y be the distance the fork has fallen when its speed is us, use a constant-acceleration equation to relate y and us:

gyvu 220

2s +=

or, because v0 = 0, gyu 22

s =

Solve for y: g

uy2

2s=

Substitute numerical values and evaluate y:

( )( ) m58.92

m/s9.812m/s34

2

2

==y

Using a constant-acceleration equation, relate the speed of the falling tuning fork to its time of fall t1:

10s gtvu +=

or, because v0 = 0, 1s gtu =

Solve for and evaluate t1: s3.466

m/s9.81m/s34.0

2s

1 ===gu

t

Chapter 15

1214

Using the relationship between distance traveled, time and average speed, find the time t2 for the sound to travel back to the student:

s0.173m/s340

m58.922 ===

vyt

Substitute in equation (1) and evaluate d:

( )( )m65.0

s0.173s3.466m/s9.81 2221

=

+=d

99 •• Picture the Problem The angle θ of the Cerenkov shock wave is related to the speed of light in water v and the speed of light in a vacuum c according to .sin cv=θ

Relate the speed of light in water v to the angle of the Cerenkov cone:

cv

=θsin

Solve for v: θsincv =


( )m/s1025.2

75.48sinm/s10998.28

8

×=

°×=v

General Problems 100 • Picture the Problem The equation of a wave traveling in the positive x direction is of the form ( ) )( vtxfx,ty −= and that of a wave traveling in the negative x direction is ( ) )( vtxfx,ty += .

Wave Motion

1215

(a) The pulse at t = 0 shown below was plotted using a spreadsheet program:

0.000

0.005

0.010

0.015

0.020

0.025

0.030

0.035

-6 -4 -2 0 2 4 6

x (m)

y(x

,0) (

m)

(b) The wave function must be of the form:

( ) ( )[ ]txfvtxfx,ty m/s10)( −=−=

because v = 10 m/s

Replace x with ( )tx m/s10− to obtain: ( )

( ) ( )[ ]22

3

m/s10m00.2m12.0

txx,ty

−+=

(c) The wave function must be of the form:

( ) ( )[ ]txfvtxfx,ty m/s10)( +=+=

because v = 10 m/s

Replace x with ( )tx m/s10+ to obtain: ( )

( ) ( )[ ]22

3

m/s10m00.2m12.0,

txtxy

++=

101 • Picture the Problem Let the subscript 1 refer to the initial situation−a tension of 800 N and a wavelength of 24 cm. Let the subscript 2 refer to the conditions that the tension is 600 N and the wavelength unknown. We can express the wavelengths of the waves on the wire in terms of the two tensions in the wire and then eliminate the constant frequency by expressing the ratio of the two wavelengths. Finally, we can solve this equation for λ2. Express the wavelength in terms of the frequency and speed of the f

v=λ

Chapter 15

1216

wave: Express the speed of the wave as a function of the tension in the wire:

µTv =


µλ T

f1

=

Express the wavelength when the tension in the wire is 600 N:

µλ 2

21 Tf

=

Express the wavelength when the tension in the wire is 800 N:

µλ 1

11 Tf

=

Divide the first of these equations by the second and solve for λ2:

1

212 T

Tλλ =

Substitute numerical values and evaluate λ2:

( ) cm20.8N800N600cm242 ==λ

102 • Picture the Problem Let m represent the mass of the rubber tubing whose length is L. We can express the travel time for the pulse t in terms of the separation of the post and the pulley and its speed. The speed of the pulse, in turn, can be found from the tension in the rubber tubing and its linear density. Express the time required for the pulse to travel the length of the tubing in terms of its speed and the length of the rubber tubing:

vLt =

Relate the speed of the pulses to the tension in the tubing:

mFL

LmFFv ===

µ

Substitute for v and simplify to obtain: F

Lmt =

Substitute numerical values and evaluate t:

( )( ) s0.252N110

kg0.7m10==t

Wave Motion

1217

103 • Picture the Problem The diagram shows the boat traveling on a still lake with a speed v. A bow wave generated a time t earlier is shown at an angle of θ with the direction of the boat’s motion. We can use trigonometry to relate the speed of the bow wave to the speed of the boat. Using the diagram, relate u and v to the angle θ: u

vutvt

==θsin

Solve for v: θsinuv =


( ) m/s3.42sin20m/s10 =°=v

104 • Picture the Problem The frequencies and wavelengths of the sound waves are related to the speed of sound through f = v/λ. (a) Use f = v/λ to find f:

( ) Hz113m0.310

m/s340==f

(b) Proceed as in (a):

( ) kHz11.3m0.31.0

m/s340==f

(c) Proceed as in (a) and (b):

( ) Hz675m0.0610

m/s340==f

( ) kHz7.65m0.061.0

m/s340==f

Chapter 15

1218

105 • Picture the Problem The diagram depicts the whistle traveling in a circular path of radius r = 1 m. The stationary listener will hear the maximum frequency when the whistle is at point 1 and the minimum frequency when it is at point 2. These maximum and minimum frequencies are determined by f0 and the tangential speed us = 2π r/T. We can relate the frequencies heard at point P to the speed of the approaching whistle at point 1 and the speed of the receding whistle at point 2.

Relate the frequency heard at point P to the speed of the approaching whistle at point 1:

ss

max 11 f

vuf

−=

Use the relationship between translational velocity and angular velocity to find the speed us of the whistle:

( )

m/s18.85rev

rad2s

rev3m1s

=

⎟⎠⎞

⎜⎝⎛ ×==

πru ω

Substitute numerical values and evaluate fmax:

( )

Hz529

Hz500

m/s340m/s18.851

1max

=

−=f

Relate the frequency heard at point P to the speed of the receding whistle at point 2:

ss

min 11 f

vuf

+=

Substitute numerical values and evaluate fmin:

( )

Hz474

Hz500

m/s340m/s18.851

1min

=

+=f

106 • Picture the Problem The crest-to-crest separation of the waves is their wavelength. We can find the frequency of the waves from v = fλ. When you lift anchor and head out to sea

Wave Motion

1219

you’ll become a moving receiver and we can apply ss

rr f

uvuvf

±±

= to calculate the

frequency you’ll observe. (a) Express the frequency of the ocean waves in terms of their speed and wavelength:

λvf =0


Hz0.593m15m/s8.9

0 ==f

(b) Express the frequency of the waves in terms of their speed and the speed of a moving receiver:

( ) sfvuf rr 1+=

Substitute numerical values and evaluate fr:

( ) Hz1.59Hz0.593m/s8.9m/s151r =⎟⎟

⎠

⎞⎜⎜⎝

⎛+=f

107 •• Picture the Problem Let t be the time of travel of the lefthand pulse and the subscripts L and R refer to the pulse coming from the left and right, respectively. Because the pulse traveling from the right starts later than the pulse from the left, its travel time is t − ∆t, where ∆t = 25 ms. Both pulses travel at the same speed and the sum of the distances they travel is 12 m. Express the total distance the two pulses travel: ( )ttvvt

ddd∆−+=

+= RL

Solve for vt to obtain:

( )tvdvt ∆+= 21

The speed of the pulse is given by:

LmFFv ==

µ

Substitute for v to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛∆+= t

LmFdvt 2

1

Chapter 15

1220

Substitute numerical values and evaluate vt:

( )( ) ( ) m99.7s1025kg0.085

m12N180m12 321 =⎥

⎦

⎤⎢⎣

⎡×+= −vt

*108 •• Picture the Problem Let the frequency of the car’s horn be fs, the frequency you hear as the car approaches fr, and the frequency you hear as the car recedes fr′. We can use

ss

rr f

uvuvf

±±

= to express the frequencies heard as the car approaches and recedes and

then use these frequencies to express the fractional change in frequency as the car passes you. Express the fractional change in frequency as the car passes you:

1.0r

=∆ff

Relate the frequency heard as the car approaches to the speed of the car:

ss

r 11 f

vuf

−=

Express the frequency heard as the car recedes in terms of the speed of the car:

ss

r 11 f

vu'f

+=

Divide the second of these frequency equations by the first to obtain:

vuvu

f'f

s

s

r

r

11+−

=

and

1.0111

s

s

rr

r

r

r =+−

−=∆

=−vuvu

ff

f'f

ff

Solve us: vu

9.11.0

s =

Substitute numerical values and evaluate us:

( )

km/h4.64h

s360010km1

sm89.17

m/s3409.11.0

3

s

=

××=

=u

Wave Motion

1221

109 •• Picture the Problem The pressure amplitude can be calculated directly from

,00 vsp ρω= and the intensity from .20

221 vsI ρω= The power radiated is the intensity

times the area of the driver. (a) Relate the pressure amplitude to the displacement amplitude, angular frequency, wave velocity, and air density:

00 vsp ρω=

Substitute numerical values and evaluate p0:

( ) ( )[ ]( )( )

2

3

130

N/m1.55

m100.025m/s340

s8002kg/m1.29

=

××

=−

−πp

(b) Relate the intensity to these same quantities:

vsI 20

221 ρω=


( ) ( )[ ]( ) ( )

2

23

21321

W/m46.3

m/s340m100.025

s8002kg/m1.29

=

××

=−

−πI

(c) Express the power in terms of the intensity and the area of the driver:

IrIAP 2π==


( ) ( ) W109.0W/m3.46m1.0 22 == πP

110 •• Picture the Problem The frequency of the sound wave is related to the density of the air, displacement amplitude, and velocity by .2

02

21 vsI ρω=

Relate the intensity of the sound wave to the density of the air, displacement amplitude, velocity, and angular frequency:

vsI 20

221 ρω=

Solve for the angular frequency:

vI

s ρω 21

0

=

Chapter 15

1222

Solve for f:

vI

sf

ρπ2

21

0

=

Substitute numerical values and evaluate f: ( )

( )( )( )

kHz07.1

m/s340kg/m1.29W/m102

m1021

3

22

6

=

=−

−πf

111 •• Picture the Problem The force exerted on the plate is due to the change in momentum of the water. We can use Newton’s 2nd law in the form F = ∆p/∆t to relate F to the mass of water in a length of tube equal to vs∆t and to the speed of the water. This mass of water, in turn, is given by the product of its density and the volume of water in a length of the tube equal to vs∆t. Relate the force exerted on the plate to the change in momentum of the water:

tmv

tpF

∆∆

=∆∆

= w

Express ∆m in terms of the mass of water in a length of tube equal to vs∆t:

tAvVm ∆=∆=∆ sρρ


ws AvvF ρ=


( )( ) ( )[ ]( ) kN0.77m/s7m0.05km/s1.4kg/m01 233 == πF

112 •• Picture the Problem Let d be the horizontal distance from the soap bubble to the position of the microphone. The angle θ of the shock wave is related to the speed of sound in air u and the speed of the bullet c according to .sin vu=θ We can determine θ

from the given information and then use this angle to find d. Express d in terms of the angle of the shock wave and the distance from the soap bubble to the laboratory bench:

θtanm35.0

=d

Relate the speed of the bullet to the angle of the shock-wave cone:

vu

=θsin

Wave Motion

1223

Solve for θ to obtain: vu1sin−=θ


⎥⎦⎤

⎢⎣⎡

=−

vu

d1sintan

m35.0


cm3.26

25.1sintan

m35.01

=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

=−

uu

d

113 •• Picture the Problem The source of the problem is that it takes a finite time for the sound to travel from the front of the line of marchers to the back. We can use the given data to determine the time required for the beat to reach the marchers in the back of the column and then use this time and the speed of sound to find the length of the column. Express the length of the column in terms of the speed of sound and the time required for the beat to travel the length of the column:

tvL ∆=

Calculate the time for the sound to travel the length of the column:

s6.0min100

1==∆t

Substitute and evaluate L: ( )( ) m204s0.6m/s340 ==L

114 •• Picture the Problem The interval between the arrival times of the echo pulses heard by the bat is the reciprocal of the frequency of the reflected pulses. We can use

ss

rr f

uvuvf

±±

= to relate the frequency of the reflected pulses to the speed of the bat and

the frequency it emits. Relate the interval between the arrival times of the echo pulses heard by the bat to frequency of the reflected pulses:

r

1f

t =∆

Relate the frequency of the pulses received by the bat to its speed and the frequency it emits:

ss

rr 1

1 fvuvuf

−+

=

Chapter 15

1224

Substitute to obtain: ( ) sr

s

11

fvuvut

+−

=∆


( )ms6.11

s80m/s340

m/s121

m/s340m/s121

1

=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=∆

−

t

*115 •• Picture the Problem Let d be the distance to the moon, h be the height of earth's atmosphere, and v be the speed of light in earth’s atmosphere. We can express d ′, the distance measured when the earth’s atmosphere is ignored, in terms of the time for a pulse of light to make a round-trip from the earth to the moon and solve this equation for the length of correction d ′ − d. Express the roundtrip time for a pulse of light to reach the moon and return: c

hdvh

ttt

−+=

+=

22

atmosphere searth' ofout atmosphere searth'

Express the "measured" distance d ′ when we do not account for the atmosphere: hdh

vc

chd

vhcctd'

−+=

⎟⎠⎞

⎜⎝⎛ −

+== 2221

21

Solve for the length of correction d ′ − d:

⎟⎠⎞

⎜⎝⎛ −=− 1

vchdd'

Substitute numerical values and evaluate d ′ − d:

( )

cm0.24

199997.0

km8

=

⎟⎠⎞

⎜⎝⎛ −=−

ccdd'

Remarks: This is larger than the accuracy of the measurements, which is about 3 to 4 cm. 116 •• Picture the Problem The frequency of the waves on the wire is the same as the frequency of the tuning fork and their period is the reciprocal of the frequency. We can find the speed of the waves from the tension in the wire and its linear density. The wavelength can be determined from the frequency and the speed of the waves and the wave number from its definition. The general form of the wave function for waves on a

Wave Motion

1225

wire is ( ) ( )tkxAx,ty ω±= sin , so, once we know k and ω, because A is given, we can

write a suitable wave function for the waves on this wire. The maximum speed and acceleration of a point on the wire can be found from the angular frequency and amplitude of the waves. Finally, we can use vAP 22

21

av µω= to find the average rate at

which energy must be supplied to the tuning fork to keep it oscillating with a steady amplitude. (a) The frequency of the waves on the wire is the same as the frequency of the tuning fork:

Hz400=f

The period of the waves on the wire is the reciprocal of their frequency:

ms2.50s400

111 === −f

T

(b) Relate the speed of the waves to the tension in the wire and its linear density:

m/s316kg/m0.01

kN1===

µFv

(c) Use the relationship between the wavelength, speed and frequency of a wave to find λ:

cm79.0s400m/s316

1 === −fvλ

Using its definition, express and evaluate the wave number:

12 m95.7

m107922 −

− =×

==π

λπk

(d) Determine the angular frequency of the waves:

( ) 131 s102.51s40022 −− ×=== ππω f

Substitute for A, k, and ω in the general form of the wave function to obtain:

( ) ( ) ( )[ ( ) ]txx,ty 131 s1051.2m95.7sinmm50.0 −− ×−=

(e) Relate the maximum speed of a point on the wire to the amplitude of the waves and the angular frequency of the tuning fork:

( )( )m/s1.26

s102.51m100.5 133

max

=

××=

=−−

ωAv

Chapter 15

1226

Express the maximum acceleration of a point on the wire in terms of the amplitude of the waves and the angular frequency of the tuning fork:

( )( )23

2133

2max

m/s1015.3

s102.51m100.5

×=

××=

=−−

ωAa

(f) Express the average power required to keep the tuning fork oscillating at a steady amplitude in terms of the linear density of the wire, the amplitude of its vibrations, and the speed of the waves on the wire:

vAP 2221

av µω=


( )( ) ( ) ( ) W9.24m/s316m100.5s102.51kg/m0.1 2321321

av =××= −−P

117 ••• Picture the Problem Because the chain is rolling at high speed we can neglect the effect of gravity. The diagram shows a small portion of the chain. We’ll assume that the angle θ is small even though it shown as a large angle in the diagram. Let ∆m be the mass of the segment of the chain shown. We’ll apply Newton’s 2nd law to the segment in order to relate the tension in the chain to its linear density and speed.

(a) Apply ∑ =RvmF

2

radial to a

segment of the chain whose mass is ∆m:

RvmF

20

net ∆=

Express ∆m in terms of µ, θ, and R: θµµ Rdm ==∆ l

Express Fnet in terms of T and θ : θ21

net sin2FF =


021sin2 vF µθθ =

Wave Motion

1227

Solve for F: θ

µθ

21

20

sin2vF =

Apply the small angle approximation

:sin 21

21 θθ ≈ ( )

20

21

20

2vvF µ

θµθ

==

(b) The wave speed is the same as the speed at which the chain is moving:

µFv =0

(c) .360 through travelspulse thechain, on thepoint fixed a respect toWith

chain. theof speed theas same theischain thealong speed its becauseposition same at the remains pulse therest,at observer an by seen As

°

118 ••• Picture the Problem Let ∆m represent the mass of the segment of length ∆x = 1 mm. We can find the wave speed from the given data for the tension in the rope and its linear density. The wavelength can be found from v = fλ. We’ll use the definition of linear momentum to find the maximum transverse linear momentum of the 1-mm segment and apply Newton’s 2nd law to the segment to find the maximum net force on it. (a) Find the wave speed from the tension and linear density:

m/s10.0kg/m0.1

N10===

µFv

(b) Express the wavelength in terms of the speed and frequency of the wave:

m2.00s5m/s10

1 === −fvλ

(c) Relate the maximum transverse linear momentum of the 1-mm segment to the maximum transverse speed of the wave:

xAfxAmvp ∆=∆=∆= µπωµ 2maxmax

Substitute numerical values and evaluate pmax:

( )( )( )( )

m/skg1026.1

m0.04m101

kg/m0.1s52

4

3

1max

⋅×=

××

=

−

−

−πp

Chapter 15

1228

(d) Apply ∑ =rvmF

2

radial to the

1-mm segment and simplify to obtain:

maxmax

2222

max

2 fpp

xAA

AxAvmF

πω

ωµωµ

==

∆=∆=∆=

Substitute numerical values and evaluate Fmax:

( )( )mN96.3

m/skg101.26s52 41max

=

⋅×= −−πF

*119 ••• Picture the Problem We can relate the speed of the pulse to the tension in the rope and its linear density. Because the rope hangs vertically, the tension in it varies linearly with the distance from its bottom. Once we’ve established the result in part (a), we can integrate the resulting velocity equation to find the time for the pulse to travel the length of the rope and then double this time to get the round-trip time. (a) Relate the speed of transverse waves to tension and linear density:

µFv =

Express the force acting on a segment of the rope of length y:

ygmgF µ==

Substitute to obtain: gyygv ==

µµ

(b) Because the speed of the pulse varies with the distance from the bottom of the rope, express v as dy/dt and solve for dt:

gydtdy

= and y

dyg

dt 1=

Integrate the left side of the equation from 0 to t and the right side from 0 to 3 m:

∫∫ =m3

00

1'y

dyg

dtt

and

( )s106.1

m/s81.9

m3221

2

m3

0

=

== yg

t

The time for the pulse to make the round-trip is:

( ) s21.2s106.122trip-round === tt

Wave Motion

1229

120 ••• Picture the Problem We can follow the step-by-step instructions outlined above to obtain the given expressions for ∆U.

(a) Express the potential energy of a segment of the string:

( )xFU ∆−∆=∆ l

For ∆y/∆x << 1: ( )[ ]2211 xyx ∆∆+∆=∆l

and ( )[ ]

( ) xxy

xxyxx

∆∆∆=

∆−∆∆+∆=∆−∆2

21

2211l

Substitute to obtain: ( )[ ]

( ) xxyF

xyFU

∆∆∆=

∆∆=∆2

21

221

(b) Differentiate ( ) ( )tkxAx,ty ω−= sin to obtain:

( )tkxkAdxdy ω−= cos

Approximate ∆y/∆x by dy/dx and substitute in our result from part (a):

( )( )( )tkxxkFA

xtkxkAFU

ω

ω

−∆=

∆−=∆222

21

221

cos

cos

Chapter 15

1230

1231

Chapter 16 Superposition and Standing Waves Conceptual Problems *1 •• Picture the Problem We can use the speeds of the pulses to determine their positions at the given times.

2 •• Picture the Problem We can use the speeds of the pulses to determine their positions at the given times.

3 • Determine the Concept Beats are a consequence of the alternating constructive and destructive interference of waves due to slightly different frequencies. The amplitudes of the waves play no role in producing the beats. correct. is )(c

4 • (a) True. The harmonics for a string fixed at both ends are integral multiples of the frequency of the fundamental mode (first harmonic). (b) True. The harmonics for a string fixed at both ends are integral multiples of the

Chapter 16

1232

frequency of the fundamental mode (first harmonic). (c) True. If l is the length of the pipe and v the speed of sound, the excited harmonics are

given byl4

vnfn = , where n = 1, 3, 5…

5 •• Determine the Concept Standing waves are the consequence of the constructive interference of waves that have the same amplitude and frequency but are traveling in opposite directions. correct. is )(b

*6 • Determine the Concept Our ears and brain find frequencies which are small-integer multiples of one another pleasing when played in combination. In particular, the ear hears frequencies related by a factor of 2 (one octave) as identical. Thus, a violin sounds much more "musical" than the sound of a drum. 7 • Picture the Problem The first harmonic displacement-wave pattern in an organ pipe open at both ends and vibrating in its fundamental mode is represented in part (a) of the diagram. Part (b) of the diagram shows the wave pattern corresponding to the fundamental frequency for a pipe of the same length L that is closed at one end. Letting unprimed quantities refer to the open pipe and primed quantities refer to the closed pipe, we can relate the wavelength and, hence, the frequency of the fundamental modes using v = fλ. Express the frequency of the first harmonic in the open pipe in terms of the speed and wavelength of the waves:

11 λ

vf =

Relate the length of the open pipe to the wavelength of the fundamental mode:

L21 =λ

Superposition and Standing Waves

1233

Substitute to obtain: Lvf

21 =

Express the frequency of the first harmonic in the closed pipe in terms of the speed and wavelength of the waves:

'v'f1

1 λ=

Relate the length of the closed pipe to the wavelength of the fundamental mode:

L' 41 =λ

Substitute to obtain: 11 2

122

14

fLv

Lv'f =⎟

⎠⎞

⎜⎝⎛==

Substitute numerical values and evaluate 'f1 :

( ) Hz200Hz40021

1 =='f

and correct. is )(a

8 •• Picture the Problem The frequency of the fundamental mode of vibration is directly proportional to the speed of waves on the string and inversely proportional to the wavelength which, in turn, is directly proportional to the length of the string. By expressing the fundamental frequency in terms of the length L of the string and the tension F in it we can examine the various changes in lengths and tension to determine which would halve it. Express the dependence of the frequency of the fundamental mode of vibration of the string on its wavelength:

11 λ

vf =

Relate the length of the string to the wavelength of the fundamental mode:

L21 =λ

Substitute to obtain: Lvf

21 =

Express the dependence of the speed of waves on the string on the tension µ

Fv =

Chapter 16

1234

in the string: Substitute to obtain:

µF

Lf

21

1 =

(a) Doubling the tension and the length would increase the frequency by a factor of 22 .

(b) Halving the tension and keeping the length fixed would decrease the frequency by a factor of 21 .

(c) Keeping the tension fixed and halving the length would double the frequency.

correct. is )(c

9 •• Determine the Concept We can relate the resonant frequencies of an organ pipe to the speed of sound in air and the speed of sound to the absolute temperature. Express the dependence of the resonant frequencies on the speed of sound:

λvf =

Relate the speed of sound to the temperature of the air:

MRTv γ

=

where γ and R are constants, M is the molar mass of the gas (air), and T is the absolute temperature.

Substitute to obtain: M

RTf γλ1

=

s.frequencieresonant theincreases re temperatu theincreasing , Because Tv ∝

*10 • Determine the Concept Because the two waves move independently, neither impedes the progress of the other.


1235

11 • Determine the Concept No; the wavelength of a wave is related to its frequency and speed of propagation (λ = v/f). The frequency of the plucked string will be the same as the wave it produces in air, but the speeds of the waves depend on the media in which they are propagating. Because the velocities of propagation differ, the wavelengths will not be the same. 12 • Determine the Concept No; when averaged over a region in space including one or more wavelengths, the energy is unchanged. 13 • Determine the Concept When the edges of the glass vibrate, sound waves are produced in the air in the glass. The resonance frequency of the air columns depends on the length of the air column, which depends on how much water is in the glass. 14 •• Picture the Problem We can use v = fλ to relate the frequency of the sound waves in the organ pipes to the speed of sound in air, nitrogen, and helium. We can use

MRTv γ= to relate the speed of sound, and hence its frequency, to the properties of

the three gases. Express the frequency of a given note as a function of its wavelength and the speed of sound:

λvf =

Relate the speed of sound to the absolute temperature and the molar mass of the gas used in the organ:

MRTv γ

=

where γ depends on the kind of gas, R is a constant, T is the absolute temperature, and M is the molar mass.


MRTf γ

λ1

=

For air in the organ pipes we have:

air

airair

1M

RTf γλ

= (1)

When nitrogen is in the organ pipes:

2

2

2N

NN

1M

RTf

γλ

= (2)

Chapter 16

1236

Express the ratio of equation (2) to equation (1) and solve for

2Nf : 2

22

N

air

air

N

air

N

MM

ff

γγ

=

and

2

2

2N

air

air

NairN M

Mffγγ

=

Because airN2

γγ = and 2Nair MM > : airN2

ff >

i.e., pipe.organ each for increase willf

If helium were used, we’d have:

He

air

air

HeairHe M

Mffγγ

=

Because airHe γγ > and Heair MM >> : airHe ff >>

i.e., .pronounced

moreeven be leffect wil the

*15 •• Determine the Concept Increasing the tension on a piano wire increases the speed of the waves. The wavelength of these waves is determined by the length of the wire. Because the speed of the waves is the product of their wavelength and frequency, the wavelength remains the same and the frequency increases. correct. is )(b

16 •• Determine the Concept If connected properly, the speakers will oscillate in phase and interfere constructively. If connected incorrectly, they interfere destructively. It would be difficult to detect the interference if the wavelength is short, less than the distance between the ears of the observer. Thus, one should use bass notes of low frequency and long wavelength. 17 •• Determine the Concept The pitch is determined mostly by the resonant cavity of the mouth; the frequency of sounds he makes is directly proportional to their speed. Because vHe > vair (see Equation 15-5), the resonance frequency is higher if helium is the gas in the cavity.


1237

*18 •• Determine the Concept The light is being projected up from underneath the silk, so you will see light where there is a gap and darkness where two threads overlap. Because the two weaves have almost the same spatial period but not exactly identical (because the two are stretched unequally), there will be places where, for large sections of the cloth, the two weaves overlap in phase, leading to brightness, and large sections where the two overlap 90° out of phase (i.e., thread on gap and vice versa) leading to darkness. This is exactly the same idea as in the interference of two waves. Estimation and Approximation 19 •• Determine the Concept Pianos are tuned by ringing the tuning fork and the piano note simultaneously and tuning the piano string until the beats are far apart; i.e., the time between beats is very long. If we assume that 2 s is the maximum detectable period for the beats, then one should be able to tune the piano string to at least 0.5 Hz. *20 • Picture the Problem We can use v = f1λ1 to express the resonance frequencies in the

organ pipes in terms of their wavelengths and ... 3, 2, 1, ,2

== nnL nλto relate the length

of the pipes to the resonance wavelengths. (a) Relate the fundamental frequency of the pipe to its wavelength and the speed of sound:

11 λ

vf =

Express the condition for constructive interference in a pipe that is open at both ends:

... 3, 2, 1, ,2

== nnL nλ (1)

Solve for λ1: L21 =λ

Substitute and evaluate f1:

( ) kHz2.27m107.52

m/s3402 21 =

×== −L

vf

(b) Relate the resonance frequencies of the pipe to their wavelengths and the speed of sound:

nn

vfλ

=

Solve equation (2) for λn: nL

n2

=λ

Chapter 16

1238


( )kHz27.2m107.52

m/s3402 2

n

nLvnfn

=×

== −

Set fn = 20 kHz and evaluate n: 81.8

kHz2.27kHz20

==n

hearing. goodh very person wit aby heard bemight harmonicninth The audible. as defined range e within this harmoniceighth The

21 •• Picture the Problem Assume a pipe length of 5 m and apply the standing-wave resonance frequencies condition for a pipe that is open at both ends (the same conditions hold for a string that is fixed at both ends). Relate the resonance frequencies for a pipe open at both ends to the length of the pipe:

... 3, 2, 1, ,2

== nLvnfn

Evaluate this expression for n = 1: ( ) Hz34.0

m52m/s340

1 ==f

Express the dependence of the speed of sound in a gas on the temperature:

MRTv γ

=

where γ and R are constants, M is the molar mass, and T is the absolute temperature.

summer. in thehigher somewhat be willfrequency the Because ,Tv ∝

Superposition and Interference 22 • Picture the Problem We can use δ2

10 cos2yA = to find the amplitude of the resultant

wave. (a) Evaluate the amplitude of the resultant wave when δ = π/6:

( )

cm86.3

621cosm02.02cos2 2

10

=

⎟⎠⎞

⎜⎝⎛==

πδyA


1239

(b) Proceed as in (a) with δ = π/3:

( )

cm46.3

321cosm02.02cos2 2

10

=

⎟⎠⎞

⎜⎝⎛==

πδyA

23 • Picture the Problem We can use δ2

10 cos2yA = to find the amplitude of the resultant

wave. Evaluate the amplitude of the resultant wave when δ = π/2:

( )

cm07.7

221cosm05.02cos2 2

10

=

⎟⎠⎞

⎜⎝⎛==

πδyA

*24 • Picture the Problem The phase shift in the waves generated by these two sources is due to their separation of λ/3. We can find the phase difference due to the path difference

fromλ

πδ x∆= 2 and then the amplitude of the resultant wave from δ2

10 cos2yA = .

Evaluate the phase difference δ: π

λλπ

λπδ

32322 ==

∆=

x

Find the amplitude of the resultant wave:

AA

AyA

==

⎟⎠⎞

⎜⎝⎛==

3cos2

32

21cos2cos2 2

10res

π

πδ

25 • Picture the Problem The phase shift in the waves generated by these two sources is due to a path difference ∆x = 5.85 m – 5.00 m = 0.85 m. We can find the phase difference due

to this path difference from λ

πδ x∆= 2 and then the amplitude of the resultant wave

from .cos2 21

0 δyA =

(a) Find the phase difference due to the path difference:

λπδ x∆

= 2

Calculate the wavelength of the sound waves:

m3.4s100m/s340

1 === −fvλ

Chapter 16

1240

Substitute and evaluate δ: °=== 0.90rad2m3.4

m0.852 ππδ

(b) Relate the amplitude of the resultant wave to the amplitudes of the interfering waves and the phase difference between them:

A

AyA

2

221cos2cos2 2

10

=

⎟⎠⎞

⎜⎝⎛==

πδ

*26 • Picture the Problem The diagram is shown below. Lines of constructive interference are shown for path differences of 0, λ, 2λ, and 3λ.

27 • Picture the Problem The intensity at the point of interest is dependent on whether the speakers are coherent and on the total phase difference in the waves arriving at the given

point. We can use λ

πδ x∆= 2 to determine the phase difference δ, δ2

10 cos2 pA = to

find the amplitude of the resultant wave, and the fact that the intensity I is proportional to the square of the amplitude to find the intensity at P for the given conditions. (a) Find the phase difference δ:

πλλπδ == 2

1

2


0cos2 21

0 == πpA

Because the intensity is proportional to A2:

0=I

(b) The sources are incoherent and 02II =


1241

the intensities add: (c) Express the total phase difference:

π

ππλ

ππ

δδδ

22122

differencepathsourcestot

=

⎟⎠⎞

⎜⎝⎛+=

∆+=

+=

x


( ) 021

0 22cos2 ppA == π


( )002

0

20

020

2

42 IIppI

pAI ===

28 • Picture the Problem The intensity at the point of interest is dependent on whether the speakers are coherent and on the total phase difference in the waves arriving at the given

point. We can use λ

πδ x∆= 2 to determine the phase difference δ, δ2

10 cos2 pA = to

find the amplitude of the resultant wave, and the fact that the intensity I is proportional to the square of the amplitude to find the intensity at P for the given conditions. (a) Find the phase difference δ:

πλλπδ 22 ==


( ) 021

0 22cos2 ppA == π


( )002

0

20

020

2

42 IIppI

pAI ===

(b) The sources are incoherent and the intensities add:

02II =

(c) Express the total phase difference:

πλλππ

λππ

δδδ

3

22

differencepathsourcestot

=

⎟⎠⎞

⎜⎝⎛+=

∆+=

+=

x

Find the amplitude of the resultant ( ) 03cos2 2

10 == πpA

Chapter 16

1242

wave: Because the intensity is proportional to A2:

0=I

29 • Picture the Problem Let P be the point located a distance r1 from speaker 1 and a distance r2 from speaker 2. If the sound at point P is to be either a maximum or a minimum, the difference in the distances to the speakers will have to be such that this difference compensates for the 90° out-of-phase condition of the speakers. (a) Express the phase shift due to the speakers in terms of a path difference:

λλλδ41sources

36090

360=

°°

=°

=∆r

Express the condition that 12 rr − must

satisfy in order to compensate for this path difference:

λ41

12 =− rr

(b) In this case, the smallest difference in path is again λ/4, but now:

λ41

21 =− rr

*30 •• Picture the Problem The drawing shows a generic point P located a distance r1 from source S1 and a distance r2 from source S2. The sources are separated by a distance d and we’re given that d < λ/2. Because the condition for destructive interference is that δ = nπ where n = 1, 2, 3,..., we’ll show that, with d < λ/2, this condition cannot be satisfied.

Relate the phase shift to the path difference and the wavelength of the sound:

λπδ r∆

= 2

Relate ∆r to d and θ: ddr ≤<∆ θsin

Substitute to obtain: λ

πλ

θπδ dd 2sin2 ≤<


1243

Because 2λ<d :

π

λλπδ =<

22

Express the condition for destructive interference:

πδ n= where n = 1, 2, 3,…

direction.any in ceinterferen edestructiv complete no is there Because ,πδ <

31 •• Picture the Problem Let the positive x direction be the direction of propagation of the wave. We can express the phase difference in terms of the separation of the two points and the wavelength of the wave and solve for λ. In part (b) we can find the phase difference by relating the time between displacements to the period of the wave. I in part (c) we can use the relationship between the speed, frequency, and wavelength of a wave to find its velocity. (a) Relate the phase difference to the wavelength of the wave:

λπδ x∆

= 2

Solve for and evaluate λ: cm0.606

cm522 ==∆

=π

πδ

πλ x

(b) Express and evaluate the period of the wave:

ms25s40

111 === −f

T

Relate the time between the two displacements to the period of the wave:

T51ms5 =

Express the phase difference corresponding to one-fifth of a period:

52πδ =

(c) Express the wave speed in terms of its frequency and wavelength:

( )( ) m/s0.24m0.6s40 1 === −λfv

32 •• Picture the Problem Assume a distance of about 20 cm between your ears. When you rotate your head through 90°, you introduce a path difference of 20 cm. We can apply the equation for the phase difference due to a path difference to determine the change in phase between the sounds received by your ears as you rotate your head through 90°.

Chapter 16

1244

Express the phase difference due to the rotation of your head through 90°:

λπδ cm202=

Find the wavelength of the sound waves:

cm50s680m/s340

1 === −fvλ

Substitute to obtain: rad8.0

cm50cm202 ππδ ==

33 •• Picture the Problem Because the sound intensity diminishes as the observer moves, parallel to a line through the sources, away from her initial position, we can conclude that her initial position is one at which there is constructive interference of the sound coming from the two sources. We can apply the condition for constructive interference to relate the wavelength of the sound to the path difference at her initial position and the relationship between the velocity, frequency, and wavelength of the waves to express this path difference in terms of the frequency of the sources. Express the condition for constructive interference at (40 m, 0):

...,3,2,1, ==∆ nnr λ (1)

Express the path difference ∆r:

AB rrr −=∆

Using the Pythagorean theorem, find rB:

( ) ( )22B m2.4m40 +=r

Substitute for rB and evaluate ∆r: ( ) ( )m0.07194

m40m2.4m40 22

=

−+=∆r

Substitute in equation (1) and solve for λ: n

m07194.0=λ

Using v = fλ, express f in terms of λ and n:

( )n

nvnfn

Hz4726m0.07194

m/s340m07194.0

=

==

Evaluate f for n = 1 and 2:

Hz47261 =f and Hz45292 =f


1245

34 •• Picture the Problem Because the sound intensity increases as the observer moves, parallel to a line through the sources, away from her initial position, we can conclude that her initial position is one at which there is destructive interference of the sound coming from the two sources. We can apply the condition for destructive interference to relate the wavelength of the sound to the path difference at her initial position and the relationship between the velocity, frequency, and wavelength of the waves to express this path difference in terms of the frequency of the sources. Express the condition for destructive interference at (40 m, 0):

...,5,3,1,2

==∆ nnr λ (1)

Express the path difference ∆r:

AB rrr −=∆

Using the Pythagorean theorem, find rB:

( ) ( )22B m2.4m40 +=r

Substitute for rB and evaluate ∆r: ( ) ( )m0.07194

m40m2.4m40 22

=

−+=∆r

Substitute in equation (1) and solve for λ:

( )nn

m1439.0m07194.02==λ

Using v = fλ, express f in terms of λ:

( )n

nvnfn

Hz3632m0.1439

m/s340m1439.0

=

==

Evaluate f for n = 1 and 3:

Hz36321 =f

and Hz70893 =f

*35 •• Picture the Problem We can use the trigonometric identity

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +

=+2

cos2

cos2coscos BABABA to derive the expression given in (a) and the

speed of the envelope can be found from the second factor in this expression; i.e., from ( ) ( )[ ]txk 2/2/cos ω∆−∆ . (a) Express the amplitude of the resultant wave function y(x,t):

( )( ( ))txktxkAtxy 2211 coscos),( ωω −+−=

Chapter 16

1246

Use the trigonometric identity ⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +

=+2

cos2

cos2coscos BABABA to obtain:

⎥⎦

⎤⎟⎠⎞

⎜⎝⎛ −

+−

⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +

−+

=

⎥⎦⎤+−−

⎢⎣⎡ −+−

=

txkktxkkA

txktxktxktxkAx,ty

22cos

22cos2

2cos

2cos2)(

12212121

22112211

ωωωω

ωωωω

Substitute ωave = (ω1 + ω2)/2, kave = (k1 + k2)/2, ∆ω = ω1 - ω2 and ∆k = k1 − k2 to obtain:

( )[ ⎥⎦

⎤⎟⎠⎞

⎜⎝⎛ ∆

−∆

−= txktxkAx,ty22

coscos2)( aveaveωω

(b) A spreadsheet program to calculate y(x,t) between 0 m and 50 m at t = 0, 0.5 s, and 1 s follows. The constants and cell formulas used are shown in the table.

Cell Content/Formula Algebraic Form B11 B10+0.25 x + ∆x C10 COS($B$3*B10−$B$5*$C$9)

+ COS($B$4*B10−$B$6*$C$9) ( )0x,y

D10 COS($B$3*B10-$B$5*$D$9) + COS($B$4*B10−$B$6*$D$9)

( )s5.0x,y

E10 COS($B$3*B10−$B$5*$E$9) + COS($B$4*B10−$B$6*$E$9)

( )s1x,y

A B C D E 1 2 3 k1= 1 m−1 4 k2= 0.8 m−1 5 w1= 1 rad/s 6 w2= 0.9 rad/s 7 x y(x,0) y(x,0.5 s) y(x,1 s)8 (m) 9 0.000 2.000 4.000

10 0.00 2.000 −0.643 −1.550 11 0.25 1.949 −0.207 −1.787 12 0.50 1.799 0.241 −1.935 13 0.75 1.557 0.678 −1.984 14 1.00 1.237 1.081 −1.932

206 49.00 0.370 −0.037 0.021 207 49.25 0.397 0.003 −0.024 208 49.50 0.397 0.065 −0.075 209 49.75 0.364 0.145 −0.124 210 50.00 0.298 0.237 −0.164


1247

The solid line is the graph of y(x,0), the dashed line that of y(x,0.5 s), and the dotted line is the graph of y(x,1 s).

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

0 5 10 15 20 25 30 35 40 45 50

x (m)

y (x ,

t )

f(x,0)f(x,0.5 s)f(x,1 s)

(c) Express the speed of the envelope: 21

21envelope kkk

v−−

=∆∆

=ωωω

Substitute numerical values and evaluate venvelope:

m/s500.0m8.0m1rad/s9.0rad/s1

11envelope =−−

= −−v

36 •• Picture the Problem The diagram shows the two sources separated by a distance d and the path difference ∆s. Because the lines from the sources to the distant point are approximately parallel, the triangle shown in the diagram is approximately a right triangle and we can use trigonometry to express ∆s in terms of d and θ. In the second part of the problem, we can apply a small-angle approximation to the larger triangle shown in Figure 16-29 to relate ym to D and θ and then use the condition for constructive interference to relate ym to D, λ, and d.

(a) Using the diagram, relate ∆s to the separation of the sources and the angle θ:

ds∆

≈θsin and θsinds ≈∆

(b) For θ << 1, we can approximate θtands ≈∆

Chapter 16

1248

sinθ with tanθ to obtain: Referring to Figure 16-29, express tanθ in terms of y and D:

Dym≈θtan

Substitute to obtain: D

dys m≈∆

Express the condition on the phase difference for constructive interference:

...,3,2,122 ==∆

= mm,s πλ

πδ

Substitute for ∆s: ...,3,2,122 == mm,Ddym π

λπ

Simplify and solve for ym:

dDmym

λ=

37 •• Picture the Problem Because a maximum is heard at 0° and the sources are in phase, we can conclude that the path difference is 0. Because the next maximum is heard at 23°, the path difference to that position must be one wavelength. We can use the result of part (a) of Problem 36 to relate the separation of the sources to the path difference and the angle θ. We’ll apply the condition for constructive interference to determine the angular locations of other points of maximum intensity in the interference pattern. Using the result of part (a) of Problem 36, express the separation of the sources in terms of ∆s and θ :

θsinsd ∆

=

Evaluate d with ∆s = λ and θ = 23°:

( ) m1.81sin23s480m/s340

23sin23sin

1 =°

=

°=

°=

−

fvd λ

Express the condition for additional intensity maxima:

λθ md m =sin

where m = 1, 2, 3, …, or

⎥⎦⎤

⎢⎣⎡= −

dm

mλθ 1sin


1249

Evaluate this expression for m = 2:

( )( )( ) °=⎥

⎦

⎤⎢⎣

⎡= −

− 5.51m1.81s480

m/s3402sin 11

2θ

Remarks: It is easy to show that, for m > 2, the inverse sine function is undefined and that, therefore, there are no additional relative maxima at angles larger than 51.5°. *38 ••• Picture the Problem Because the speakers are driven in phase and the path difference is 0 at her initial position, the listener will hear a maximum at (D, 0). As she walks along a line parallel to the y axis she will hear a minimum wherever it is true that the path difference is an odd multiple of a half wavelength. She will hear an intensity maximum wherever the path difference is an integral multiple of a wavelength. We’ll apply the condition for destructive interference in part (a) to determine the angular location of the first minimum and, in part (b), the condition for constructive interference find the angle at which she’ll hear the first maximum after the one at 0°. In part (c), we can apply the condition for constructive interference to determine the number of maxima she can hear as keeps walking parallel to the y axis. (a) Express the condition for destructive interference: 2

sin λθ md m =

where m = 1, 3, 5,…, or

⎟⎠⎞

⎜⎝⎛= −

dm

m 2sin 1 λθ

Evaluate this expression for m = 1: ( )( )

°=

⎥⎦

⎤⎢⎣

⎡=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

−−

14.8

m2s0062m/s340sin

2sin 1

111 fd

vθ

(b) Express the condition for additional intensity maxima:

λθ md m =sin

where m = 0, 1, 2, 3,…, or

⎟⎠⎞

⎜⎝⎛= −

dm

mλθ 1sin

Evaluate this expression for m = 1: ( )( )

°=

⎥⎦

⎤⎢⎣

⎡=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −

−−

5.16

m2s006m/s340sinsin 1

111 fd

vθ

Chapter 16

1250

(c) Express the limiting condition on sinθ :

1sin ≤=d

mmλθ

Solve for m to obtain: ( )( ) 53.3

m/s340m2s600 1

===≤−

vfddm

λ

Because m must be an integer: 3=m

39 ••• Picture the Problem Let d be the separation of the two sound sources. We can express the wavelength of the sound in terms of the d and either of the angles at which intensity maxima are heard. We can find the frequency of the sources from its relationship to the speed of the waves and their wavelengths. Using the condition for constructive interference, we can find the angles at which intensity maxima are heard. Finally, in part (d), we’ll use the condition for destructive interference to find the smallest angle for which the sound waves cancel. (a) Express the condition for constructive interference:

λθ md m =sin (1)

where m = 0, 1, 2, 3,…

Solve for λ: m

d mθλ sin=

Evaluate λ for m = 1: ( ) ( )

m279.0

rad140.0sinm2

=

=λ

(b) Express the frequency of the sound in terms of its wavelength and speed:

kHz22.1m0.279

m/s340===

λvf

(c) Solve equation (1) for θm: ( )

( )[ ]m

md

mm

1395.0sin

m2m279.0sinsin

1

11

−

−−

=

⎥⎦

⎤⎢⎣

⎡=⎟

⎠⎞

⎜⎝⎛=

λθ

The table shows the values for θ as a function of m:

m θm (rad)

3 0.432 4 0.592 5 0.772


1251

6 0.992 7 1.354 8 undefined

(d) Express the condition for destructive interference:

2sin λθ md m =

where m = 1, 3, 5,…

Solve for θm: ⎟⎠⎞

⎜⎝⎛= −

dmm 2

sin 1 λθ

Evaluate this expression for m = 1:

( ) rad0698.0m22

m0.279sin 11 =⎥

⎦

⎤⎢⎣

⎡= −θ

40 ••• Picture the Problem The total phase shift in the waves arriving at the points of interest is the sum of the phase shift due to the difference in path lengths from the two sources to a given point and the phase shift due to the sources being out of phase by 90°. From Problem 39 we know that λ = 0.279 m. Using the conditions on the path difference ∆x for constructive and destructive interference, we can find the angles at which intensity maxima are heard. Letting the subscript ″pd ″ denote ″path difference″ and the subscript ″s″ the ″sources″, express the total phase shift δ:

42spd

πλ

πδδδ +∆

=+=x

where ∆x is the path difference between the two sources and the points at which constructive or destructive interference is heard.

Express the condition for constructive interference:

...,6424

2 ππππλ

πδ ,,x=+

∆=

Solve for ∆x to obtain:

( )λλλλ8

18...823

815

87 −

==∆m,,,x

where m = 1, 2, 3,…

Relate ∆x to d to obtain:

( )csin

818 θλ dmx =

−=∆

where the ″c″ denotes constructive interference.

Chapter 16

1252

Solve for θc: ( )⎥⎦⎤

⎢⎣⎡ −

= −

dm8

18sin 1c

λθ , m = 1, 2, 3,…

The table shows the values for θc for m = 1 to 5:

m θc 1 °7.01

2 °2.15

3 °6.23

4 °1.35

5 °8.42

Express the condition for destructive interference:

...534

2 ,,,x ππππλ

πδ =+∆

=

Solve for ∆x to obtain:

( )λλλλ8

58...8

19811

83 −

==∆m,,,x

where m = 1, 2, 3,…

Letting ″d ″ denotes destructive interference, relate ∆x to d to obtain:

( )dsin

858 θλ dmx =

−=∆

Solve for θd: ( )

⎥⎦⎤

⎢⎣⎡ −

= −

dm

858sin 1

dλθ , m = 1, 2, 3,…

The table shows the values for θd for m = 1 to 5:

m θd 1 °00.3

2 °1.11

3 °3.19

4 °1.28

5 °6.37

41 ••• Picture the Problem We can calculate the required phase shift from the path difference

and the wavelength of the radio waves using λ

πδ s∆= 2 .

Express the phase delay as a function of the path difference and λ

πδ s∆= 2 (1)


1253

the wavelength of the radio waves:

Find the wavelength of the radio waves:

m15s1020

m/s10316

8

=×

×== −f

vλ

Express the path difference for the signals coming from an angle θ with the vertical:

θsinds =∆

Substitute numerical values and evaluate ∆s:

( )λλ

λ0.3152

2.315m34.73sin10m200+=

==°=∆s

Substitute in equation (1) and evaluate δ:

°=== 113rad98.1315.02λ

λπδ

Beats 42 • Picture the Problem The beat frequency is the difference between the frequency of the tuning fork and the frequency of the violin string. Let f2 = 500 Hz. (a) Express the relationship between the beat frequency of the frequencies of the two tuning forks:

Hz4Hz50012

±=∆±= fff

Solve for f2: Hz496orHz5042 =f

(b) Hz. 496

,diminished isit if Hz; 504 then increased, isfrequency beat theIf

2

2

==

ff

43 •• Picture the Problem The Doppler shift of the siren as heard by one of the drivers is given by the formula for source and receiver both moving and approaching each other ( ) ( )[ ]vuvuff /1/1sr −+= , where u is the speed of the ambulance and v is the speed of sound. (a) Express the beat frequency:

srbeat fff −= where fr is the frequency heard by either driver due to the other’s siren,

Chapter 16

1254

Express fr:

vuvu

ff−

+=

1

1sr


1

2

11

1

1

1

s

sssbeat

−=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−

+=−

−

+=

uv

f

vuvu

ff

vuvu

ff

Substitute numerical values and evaluate fbeat: ( ) Hz5.70

1m/s4.22m/s3402Hz500beat =

−=f

(b) Hz). 35 (approx.amount same by the up shiftedDoppler areambulances

both of sirens theasfrequency beat no hearsstreet on theperson The

Standing Waves *44 • Picture the Problem We can use v = fλ to relate the second-harmonic frequency to the wavelength of the standing wave for the second harmonic. Relate the speed of transverse waves on the string to their frequency and wavelength:

22λfv =

Express λ2 in terms of the length L of the string:

L=2λ

Substitute for λ2 and evaluate v: ( )( ) m/s180m3s60 12 === −Lfv

45 • Picture the Problem We can find the wavelength of this standing wave from the standing-wave condition for a string fixed at both ends and its frequency from v = f3λ3. We can use the wave function for a standing wave on a string fixed at both ends ( ( ) txkAx,ty nnnn ωcossin= )) to write the wave function for the wave described in this

problem.


1255

(a) Using the standing-wave condition for a string fixed at both ends, relate the length of the string to the wavelength of the harmonic mode in which it is vibrating:

... 3, 2, 1, ,2

== nnL nλ

Solve for λ3: ( ) m2.00m332

32

3 === Lλ

Express the frequency of the third harmonic in terms of the speed of transverse waves on the string and their wavelength:

Hz25.0m2m/s50

33 ===

λvf

(b) Write the equation for a standing wave, fixed at both ends, in its third harmonic:

( ) txkAtxy 3333 cossin, ω=

Evaluate k3: 1

33 m

m222 −=== ππ

λπk

Evaluate ω3: ( ) 11

33 s50s2522 −− === πππω f

Substitute to obtain: ( ) ( ) tkxx,ty ωcossinmm43 = where k = π m−1 and

ω = 50π s−1. 46 • Picture the Problem The first harmonic displacement-wave pattern in an organ pipe open at both ends and vibrating in its fundamental mode is represented in part (a) of the diagram. Part (b) of the diagram shows the wave pattern corresponding to the fundamental frequency for a pipe of the same length L that is closed at one end. We can relate the wavelength to the frequency of the fundamental modes using v = fλ. (a) Express the dependence of the frequency of the fundamental mode of vibration in the open pipe on its

open,1open,1 λ

vf =

Chapter 16

1256

wavelength: Relate the length of the open pipe to the wavelength of the fundamental mode:

L2open,1 =λ

Substitute and evaluate f1,open: ( ) Hz17.0

m102m/s340

2open,1 ===Lvf

(b) Express the dependence of the frequency of the fundamental mode of vibration in the closed pipe on its wavelength:

closed,1closed,1 λ

vf =

Relate the length of the closed pipe to the wavelength of the fundamental mode:

L4closed,1 =λ

Substitute to obtain: ( ) Hz50.8

m104m/s340

4closed,1 ===Lvf

47 • Picture the Problem We can find the speed of transverse waves on the wire using

µFv = and the wavelengths of any harmonic from ... 3, 2, 1, ,2

== nnL nλ. We can

use v = fλ to find the frequency of the fundamental. For a wire fixed at both ends, the higher harmonics are integer multiples of the first harmonic (fundamental). (a) Relate the speed of transverse waves on the wire to the tension in the wire and its linear density:

LmFFv ==

µ


( ) ( ) m/s521m1.4kg0.005

N968==v

(b) Using the standing-wave condition for a wire fixed at both ends, relate the length of the wire to the wavelength of the harmonic mode in which it is vibrating:

... 3, 2, 1, ,2

== nnL nλ

Solve for λ1: ( ) m2.80m4.1221 === Lλ


1257

Express the frequency of the first harmonic in terms of the speed and wavelength of the waves:

Hz861m2.80

m/s2151

1 ===λvf

(c) Because, for a wire fixed at both ends, the higher harmonics are integer multiples of the first harmonic:

( ) Hz372Hz18622 12 === ff

and ( ) Hz585Hz18633 13 === ff

48 •

Picture the Problem We can use Equation 16-13,

...,,5,3,14 1 === n,nf

Lvnfn to find

the resonance frequencies for a rope that is fixed at one end. (a) Using the resonance-frequency condition for a rope fixed at one end, relate the resonance frequencies to the speed of the waves and the length of the rope:

...,5,3,14 1 === n,nf

Lvnfn

Solve for f1: ( ) Hz1.25

m44m/s20

1 ==f

(b) harmonic. second

asupport not does system theend, onejust at fixed is rope thisBecause

(c) For the third harmonic, n = 3:

( ) Hz75.3Hz25.133 13 === ff

49 • Picture the Problem We can find the fundamental frequency of the piano wire using the general expression for the resonance frequencies of a wire fixed at both ends,

... 3, 2, 1, ,2 1 === nnf

Lvnfn , with n = 1. We can use µFv = to express the

frequencies of the fundamentals of the two wires in terms of their linear densities. Relate the fundamental frequency of the piano wire to the speed of transverse waves on it and its linear density:

Lvf

21 =

Chapter 16

1258

Express the dependence of the speed of transverse waves on the tension and linear density:

µFv =


µF

Lf

21

1 =

Doubling the linear density results in a new fundamental frequency f ′ given by:

11 21

21

21

221 fF

LF

L'f =⎟⎟

⎠

⎞⎜⎜⎝

⎛==

µµ

Substitute for f1 to obtain: ( ) Hz141Hz2002

11 =='f

*50 • Picture the Problem Because the frequency and wavelength of sounds waves are inversely proportional, the greatest length of the organ pipe corresponds to the lowest frequency in the normal hearing range. We can relate wavelengths to the length of the pipes using the expressions for the resonance frequencies for pipes that are open at both ends and open at one end. Find the wavelength of a 20-Hz note: m17

s20m/s340

1lowest

max === −fvλ

(a) Relate the length L of a closed-at-one-end organ pipe to the wavelengths of its standing waves:

... 5, 3, 1, ,4

== nnL nλ

Solve for and evaluate λ1: m4.254m17

4max ===

λL

(b) Relate the length L of an open organ pipe to the wavelengths of its standing waves:

... 3, 2, 1, ,2

== nnL nλ

Solve for and evaluate λ1: m50.82m17

2max ===

λL

51 •• Picture the Problem We can find λ and f by comparing the given wave function to the general wave function for a string fixed at both ends. The speed of the waves can then be


1259

found from v = fλ. We can find the length of the string from its fourth harmonic wavelength. (a) Using the wave function, relate k and λ:

1cm20.02 −==λπk

Solve for λ: cm4.31cm10

cm20.02

1 === − ππλ

Using the wave function, relate f and ω:

1s3002 −== fπω

Solve for f: Hz7.47

2s300 1

==−

πf

(b) Express the speed of transverse waves in terms of their frequency and wavelength:

( )( )m/s15.0

m0.314Hz47.7

=

== λfv

(c) Relate the length of the string to the wavelengths of its standing-wave patterns:

... 3, 2, 1, ,2

== nnL nλ

Solve for L when n = 4: ( ) cm62.8cm31.422 4 === λL

52 •• Picture the Problem We can find λ and f by comparing the given wave function to the general wave function for a string fixed at both ends. The speed of the waves can then be found from v = fλ. In a standing wave pattern, the nodes are separated by one-half wavelength. (a) Express the speed of the traveling waves in terms of their frequency and wavelength:

λfv =

Using the wave function, relate k and λ:

1m5.22 −==λπk

Solve for λ: m2.51m8.0

m.522

1 === − ππλ

Using the wave function, relate ω 1s5002 −== fπω

Chapter 16

1260

and f:

Solve for f: Hz6.79

2s500 1

==−

πf

Substitute to find v: ( )( ) m/s200m2.51s79.6 1 == −v

Express the amplitude of the standing wave in terms of the amplitude of the two traveling waves that result in the standing wave:

AA 2SW =

Solve for and evaluate A: cm50.22

m0.052SW ===

AA

(b) The distance between nodes is half the wavelength:

m1.262

m2.512

==λ

(c) Because there is a standing wave on the string, the shortest possible length is:

m1.262min ==λL


1261

53 •• Picture the Problem We can evaluate the wave function of Problem 52 at the given times to obtain graphs of position as a function of x. We can find the period of the motion from its frequency f and find f from its angular frequency ω. (a) The function y(x,0) is shown below:

-5

-4

-3

-2

-1

0

12

3

4

5

0.0 0.5 1.0 1.5 2.0 2.5

x (m)

y (x ,

0) (c

m)

The functions y(x,T/4) and y(x,3T/4) are shown below. Because these functions are identical, only one graph is shown.

-5

-4

-3

-2

-1

0

1

2

3

4

5

0.0 0.5 1.0 1.5 2.0 2.5

x (m)

y (x ,

T/4

) (cm

)

Chapter 16

1262

The function y(x,T/2) follows:

-5

-4

-3

-2

-1

0

1

2

3

4

5

0.0 0.5 1.0 1.5 2.0 2.5

x (m)

y (x ,

T/2

) (cm

)

(b) Express the period in terms of the frequency:

fT 1

=

Using the wave function, relate ω and f:

1s5002 −== fπω

Solve for f: Hz6.79

2s500 1

==−

πf

Substitute for f and evaluate T: ms12.6s79.6

11 == −T

(c) energy. kineticentirely is wave theofenergy the allfor 0 )(

when downwardor upwardeither moving is string theBecausex,xy =

*54 •• Picture the Problem Whether these frequencies are for a string fixed at one end only rather than for a string fixed at both ends can be decided by determining whether they are integral multiples or odd-integral multiples of a fundamental frequency. The length of the string can be found from the wave speed and the wavelength of the fundamental frequency using the standing-wave condition for a string with one end free. (a) Letting the three frequencies be represented by f'''f''f' and ,, , find

the ratio of the first two frequencies:

53

Hz125Hz75

==f''f'


1263

Find the ratio of the second and third frequencies:

75

Hz175Hz251

==f'''f''

(b) only. end oneat fixed bemust string theso harmonics,even no are There

(c) Express the resonance frequencies in terms of the fundamental frequency:

... 5, 3, 1, ,1 == nnffn

Noting that the frequencies are multiples of 25 Hz, we can conclude that:

Hz253Hz75

33

1 ===ff

(d) harmonics.seventh and fifth, third, thearethey

frequency, lfundamenta the times7 and 5, 3, are sfrequencie theBecause

(e) Express the length of the string in terms of the standing-wave condition for a string fixed at one end:

... 5, 3, 1, ,4

== nnL nλ

Using v = f1λ1, find λ1: m16s25m/s400

11

1 === −fvλ

Evaluate L for λ1 = 16 m and n = 1: m4.00

4m16

41 ===

λL

55 •• Picture the Problem The lowest resonant frequency in this closed-at-one-end tube is its fundamental frequency. This frequency is related to its wavelength through v = fminλmax. We can use the relationship between the nth harmonic and the fundamental frequency, ( ) ... 3, 2, 1, ,12 1n =+= nfnf , to find the highest frequency less than or equal

to 5000 Hz that will produce resonance. (a) Express the length of the space above the water in terms of the standing-wave condition for a closed pipe:

... 5, 3, 1, ,4

== nnL nλ

Solve for λn: ... 5, 3, 1, ,4== n

nL

nλ

Chapter 16

1264

λmax corresponds to n = 1: ( ) m4.8m1.244max === Lλ

Using v = fminλmax, find fmin: Hz70.8

m4.8m/s340

maxmin ===

λvf

(b) Express the nth harmonic in terms of the fundamental frequency (first harmonic):

( ) ... 3, 2, 1, ,12 1n =+= nfnf

To find the highest harmonic below 5000 Hz, let fn = 5000 Hz:

( )( )Hz8.7012Hz5000 += n

Solve for n (an integer) to obtain: n = 34

Evaluate f34: ( ) kHz4.89Hz70.89669 134 === ff

(c) There are 34 harmonics higher than the fundamental frequency so the total number is:

35

56 •• Picture the Problem Sound waves of frequency 460 Hz are excited in the tube, whose length L can be adjusted. Resonance occurs when the effective length of the tube Leff = L + ∆L equals ,,, λλλ 4

543

41 and so on, where λ is the wavelength of the sound.

Even though the pressure node is not exactly at the end of the tube, the wavelength can be found from the fact that the distance between water levels for successive resonances is half the wavelength. We can find the speed from λfv = and the end correction from the fact that, for the fundamental, L,LL ∆+== 14

1eff λ where L1 is the distance from the top

of the tube to the location of the first resonance. (a) Relate the speed of sound in air to its wavelength and the frequency of the tuning fork:

λfv =

Using the fact that nodes are separated by one-half wavelength, find the wavelength of the sound waves:

( )cm75

cm18.3cm8.552=

−=λ

Substitute and evaluate v: ( )( ) m/s345m0.75s460 1 == −v


1265

(b) Relate the end correction ∆L to the wavelength of the sound and effective length of the tube:

LLL

∆+=

=

1

41

eff λ

Solve for and evaluate ∆L:

( )cm0.450

cm18.3cm7541

141

=

−=−=∆ LL λ

*57 •• Picture the Problem We can use v = fλ to express the fundamental frequency of the

organ pipe in terms of the speed of sound and MRTv γ

= to relate the speed of sound and

the fundamental frequency to the absolute temperature. Express the fundamental frequency of the organ pipe in terms of the speed of sound:

λvf =

Relate the speed of sound to the temperature:

MRTv γ

=

where γ and R are constants, M is the molar mass, and T is the absolute temperature.


MRTf γ

λ1

=

Using primed quantities to represent the higher temperature, express the new frequency as a function of T:

MRT'

'f' γ

λ1

=

As we have seen, λ is proportional to the length of the pipe. For the first question, we assume the length of the pipe does not change, so λ = λ′. Then the ratio of f ′ to f is:

TT'

ff'

=

Chapter 16

1266

Solve for and evaluate f ′ with T ′ = 305 K and T = 289 K:

( )

Hz452

K289K305Hz0.440

K289K305

K289K305

=

=

== fff'

re. temperatuoft independen is pipe, theoflength theis where/ that so expand pipe thehave better to be It would LL,v

58 •• Picture the Problem We can express the wavelength of the fundamental in a pipe open at both ends in terms of the effective length of the pipe using ( )LLL ∆+== 22 effλ , where L is the physical length of the pipe and λ = v/f. Solving these equations simultaneously will lead us to an expression for L as a function of D. Express the wavelength of the fundamental in a pipe open at both ends in terms of the pipe’s effective length Leff:

( )LLL ∆+== 22 effλ

where L is its physical length.

Solve for L to obtain: DLL 3186.022

−=∆−=λλ

Express the wavelength of middle C in terms of its frequency f and the speed of sound v:

fv

=λ


Df

vL 3186.02

−=

Substitute numerical values to express L as a function of D: ( )

D

DL

3186.0m664.0

3186.0s2562

m/s3401-

−=

−=

Evaluate L for D = 1 cm:

( )cm1.66

m01.03186.0m664.0

=

−=L


( )cm2.63

m1.03186.0m664.0

=

−=L


1267


( )cm8.56

m3.03186.0m664.0

=

−=L

59 •• Picture the Problem We know that, when a string is vibrating in its fundamental mode, its ends are one-half wavelength apart. We can use v = fλ to express the fundamental frequency of the organ pipe in terms of the speed of sound and µFv = to relate the

speed of sound and the fundamental frequency to the tension in the string. We can use this relationship between f and L, the length of the string, to find the length of string when it vibrates with a frequency of 650 Hz. (a) Express the wavelength of the standing wave, vibrating in its fundamental mode, to the length L of the string:

( ) cm80cm4022 === Lλ

(b) Relate the speed of the waves combining to form the standing wave to its frequency and wavelength:

v = fλ

Express the speed of transverse waves as a function of the tension in the string:

µFv =

Substitute and solve for F to obtain: L

mfF 22λ=

where m is the mass of the string and L is its length.


( ) ( )

N480

m0.4kg101.2

m0.8s5003

221

=

×=

−−F

(c) Using v = fλ and assuming that the string is still vibrating in its fundamental mode, express its frequency in terms of its length:

Lvvf

2==

λ

Chapter 16

1268

Solve for L: f

vL2

=

Letting primed quantities refer to a second length and frequency, express L′ in terms of f ′:

f'vL'

2=

Express the ratio of L′ to L and solve for L′:

f'f

LL'

= ⇒ Lf'fL' =

Evaluate L650 Hz:

( ) cm77.30cm40Hz650Hz500Hz650Hz500

Hz500Hz650

==

= LL

bridge. scroll thefrom cm 9.23finger your place shouldYou

60 •• Picture the Problem Let f′ represent the frequencies corresponding to the A, B, C, and D notes and x(f ′) represent the distances from the end of the string that a finger must be placed to play each of these notes. Then, the distances at which the finger must be placed are given by ( ) ( ) ( )'fLfL'fx −= G .

Express the distances at which the finger must be placed in terms of the lengths of the G string and the frequencies f ′ of the A, B, C, and D notes:

( ) ( ) ( )'G fLfL'fx −= (1)

Assuming that it vibrates in its fundamental mode, express the frequency of the G string in terms of its length:

GGG 2L

vvf ==λ

Solve for LG:

GG 2 f

vL =

Letting primed quantities refer to the string lengths and frequencies of 'f

vL'2

=


1269

the A, B, C, and D notes, express L′ in terms of f ′: Express the ratio of L′ to L and solve for L′:

'ff

LL' G

G

= ⇒ GG L'f

fL' =

Evaluate L′ = L(f′) for the notes A, B, C and D to complete the table:

Note Frequency L(f ′) (Hz) (cm)

A 220 26.73 B 247 23.81 C 262 22.44 D 294 20.00

Use equation (1) to evaluate x(f ′) and complete the table to the right:

Note Frequency L(f ′) x(f ′) (Hz) (cm) (cm)

A 220 26.73 27.3

B 247 23.81 19.6

C 262 22.44 56.7

D 294 20.00 0.10

61 •• Picture the Problem We can use the fact that the resonance frequencies are multiples of the fundamental frequency to find both the fundamental frequency and the harmonic numbers corresponding to 375 Hz and 450 Hz. We can find the length of the string by relating it to the wavelength of the waves on it and the wavelength to the speed and frequency of the waves. The speed of the waves is, in turn, a function of the tension in the string and its linear density, both of which we are given. (a) Express 375 Hz as an integer multiple of the fundamental frequency of the string:

Hz3751 =nf (1)

Express 450 Hz as an integer multiple of the fundamental frequency of the string:

( ) Hz4501 1 =+ fn (2)

Solve equations (1) and (2) simultaneously for f1:

Hz0.751 =f

Chapter 16

1270

(b) Substitute in equation (1) to obtain:

5=n

sixth. andfifth theare harmonics The

(c) Express the length of the string as a function of the speed of transverse waves on it and its fundamental frequency:

122 fvL ==

λ

Express the speed of transverse waves on the string in terms of the tension in the string and its linear density:

µFv =

Substitute to obtain: µ

Ff

L12

1=

Substitute numerical values and evaluate L: ( ) m00.2

kg/m104N360

s7521

31 =×

= −−L

62 •• Picture the Problem We can use the fact that the resonance frequencies are multiples of the fundamental frequency and are expressible in terms of the speed of the waves and their wavelengths to find the harmonic numbers corresponding to wavelengths of 0.54 m and 0.48 m. We can find the length of the string by using the standing-wave condition for a string fixed at both ends. (a) Express the frequency of the nth harmonic in terms of its wavelength:

m54.01vvnf

n

==λ

Express the frequency of the (n + 1)th harmonic in terms of its wavelength:

( )m48.0

11

1vvfn

n

==++λ

Solve these equations simultaneously for n:

8=n

ninth. andeighth theare harmonics The


1271

(b) Using the standing-wave condition, both ends fixed, relate the length of the string to the wavelength of its nth harmonic:

... 3, 2, 1, ,2

== nnL nλ

Evaluate L for the eighth harmonic: m16.22

m54.08 =⎟⎠⎞

⎜⎝⎛=L

63 •• Picture the Problem The linear densities of the strings are related to the transverse wave speed and tension through .µFv = We can use v = fλ = 2fL to relate the frequencies

of the violin strings to their lengths and linear densities. (a) Relate the speed of transverse waves on a string to the tension in the string and solve for the string’s linear density:

µFv =

and

2vF

=µ

Express the dependence of the speed of the transverse waves on their frequency and wavelength:

Lffv

E

E

2== λ


E

EE 4 Lf

F=µ

Substitute numerical values and evaluate µE: ( )[ ] ( )

g/m0.574

kg/m1074.5m3.0s4405.14

N90

4

221E

=

×=

=

−

−µ

(b) Evaluate µA:

( ) ( )

g/m29.1

kg/m1029.1m3.0s4404

N90

3

221A

=

×=

=

−

−µ

Chapter 16

1272

Evaluate µD: ( ) ( )

g/m91.2

kg/m1091.2m3.0s2934

N90

3

221D

=

×=

=

−

−µ

Evaluate µG:

( ) ( )

g/m57.6

kg/m1057.6m3.0s1954

N90

3

221G

=

×=

=

−

−µ

64 •• Picture the Problem The spatial period is one-half the wavelength of the standing wave produced by the sound and its reflection. Hence we can solve ''λfc = for λ′ and use

( )[ ]cvff −= 11' to derive an expression for λ′/2 in terms of c, v, and f.

(a) Express the wavelength of the reflected sound as a function of its frequency and the speed of sound in air:

''

fc

=λ

Use the expression for the Doppler-shift in frequency when to source is in motion to obtain:

cv

ff−

=1

1'

where c is the speed of sound.


fvc

cv

fc

cv

f

cf

c

21

2

1

12'2

2'

−=⎟

⎠⎞

⎜⎝⎛ −=

−

==λ

Substitute numerical values and evaluate the spatial period of the standing wave:

( ) m318.0s5002

m/s4.22m/s340 2'

1- =−

=λ


1273

(b)

softer. andlouder get ly periodical willintensity theso ely)destructivpartially sometimes andvely constructi interfere sometimes will

wavesreflected the(i.e., resonance ofout andin movely periodical willsiren its from wavessound the wall, thecloser to moves ambulance theAs

65 •• Picture the Problem Beat frequencies are heard when the strings are vibrating with slightly different frequencies. To understand the beat frequency heard when the A and E strings are bowed simultaneously, we need to consider the harmonics of both strings. In part (c) we’ll relate the tension in the string to the frequency of its vibration and set up a proportion involving the frequencies corresponding to the two tensions that we can solve for the tension when the E string is perfectly tuned.

(a)

heard. be will2 ofbeat a,Hz) (660 If Hz. 660an greater thslightly is string E theof

frequency original theand string, E theof harmonic second theequals stringA theof harmonic third thebecausebeat a produce sounds twoThe

E

fff

∆∆ +=

(b) Because fbeat increases with increasing tension, the frequency of the E string is greater than 660 Hz. Thus the frequency of the E string is:

( )Hz661.5

Hz3Hz606 21

E

=

+=f

(c) Express the frequency of a string as a function of its tension: µλλ

Fvf 1==

When the frequency of the E string is 660 Hz we have:

µλHz6601Hz660

F=

When the frequency of the E string is 661.5 Hz we have:

µλN801Hz5.661 =

Divide the first of these equations by the second and solve for F660 Hz to obtain:

( ) N79.6N80Hz661.5

Hz6602

Hz660 =⎟⎟⎠

⎞⎜⎜⎝

⎛=F

66 •• Picture the Problem We can use the condition for constructive interference of the waves reflected from the walls in front of and behind you to relate the path difference to the

Chapter 16

1274

wavelength of the sound. We can find the wavelength of the sound from its frequency and the speed of sound in air. Express the total path difference as you walk toward the far wall of the hall:

far wallwallnear xxx ∆+∆=∆ (1)

Express the condition on the path difference for constructive interference:

xn ∆=λ where n = 1, 2, 3, … (2)

The reduction in the distance to the nearer wall as you walk a distance d is:

∆xnear wall = 2d

The increase in the distance to the farther wall as you walk a distance d is:

∆xfar wall = 2d

Substitute in equation (1) to find the total path difference as you walk a distance d:

dddx 422 =+=∆

Relate λ to f and v: fv

=λ

Substitute in equation (2) to obtain: d

fvn 4=

Solve for and evaluate d for n = 1:

( ) cm12.5s6804

m/s3404 1 === −fvd

*67 •• Picture the Problem Let the wave function for the wave traveling to the right be

( ) ( )δω −−= tkxAtxy sin,R and the wave function for the wave traveling to the left be ( ) ( )δω ++= tkxAtxy sin,L and use the identity

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +

=+2

cos2

sin2sinsin βαβαβα to show that the sum of the wave functions

can be written in the form ( ) ( )δω += tkxAtxy cossin', .


1275

Express the sum of the traveling waves of equal amplitude moving in opposite directions:

( ) ( ) ( ) ( ) ( )δωδω +++−−=+= tkxAtkxAtxytxytxy sinsin,,, LR

Use the trigonometric identity to obtain:

( )

( )δω

δωδωδωδω

−−=

⎟⎠⎞

⎜⎝⎛ −−−−−

⎟⎠⎞

⎜⎝⎛ +++−−

=

tkxA

tkxtkxtkxtkxAtxy

cossin22

cos2

sin2,

Because the cosine function is even; i.e., cos(−θ) = cosθ:

( ) ( )( )δω

δω+=+=

tkxAtkxAtxy

cossin'cossin2,

where A′ = 2A.

Thus we have: ( ) ( )δω += tkxAtxy cossin',

provided A′ = 2A. 68 •• Picture the Problem We can find ω3 and k3 from the given information and substitute to find the wave function for the 3rd harmonic. We can use the time-derivative of this expression (the transverse speed) to express the kinetic energy of a segment of mass dm and length dx of the string. Integrating this expression will give us the maximum kinetic energy of the string in terms of its mass. (a) Write the general form of the wave function for the 3rd harmonic:

( ) txkAtxy 3333 cossin, ω=

Evaluate ω3: ( ) 1133 s200s10022 −− === πππω f

Using the standing-wave condition for a string fixed at one end, relate the length of the string to its 3rd harmonic wavelength:

43 3λ

=L

and

( ) m38m2

34

34

3 === Lλ

Evaluate k3:

( )1

33 m

43

m3822 −===

ππλπk

Chapter 16

1276

Substitute numerical values and evaluate Kmax:

( ) ( )( )mmK

J/kg8.88

m03.0s200 22141

max

=

= −π


( ) ( ) ( )txtxy 113 s200cosm

43sinm03.0, −−

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛= ππ

(b) Express the kinetic energy of a segment of string of mass dm:

221

ydmvdK =

Express the mass of the segment in terms of its length dx and the linear density of the string:

dxdm µ=

Using our result in (a), evaluate vy:

( ) ( )

( )( ) ( )

( ) ( ) tx

tx

txt

vy

11

111

11

s200sinm4

3sinm/s6

s200sinm4

3sinm03.0s200

s200cosm4

3sinm03.0

−−

−−−

−−

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

∂∂

=

πππ

πππ

ππ


( ) ( ) dxtxdK µπππ2

1121 s200sinm

43sinm/s6 ⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛= −−

Express the condition on the time that dK is a maximum:

( ) 1s200sin 1 =− tπ

or

( ) ,...2

3,2

s200 1 πππ =− t

Solve for and evaluate t:

,...msms,7.5050.2

,...2

3s200

1,2s200

111

=

= −−

ππ

ππ

t


1277

Because the string’s maximum kinetic energy occurs when y(x,t) = 0:

line.straight a is string The

(c) Integrate dK from (b) over the length of the string to obtain: [ ]

[ ]22

41

041

2122

21

0

22221

0

221

max

2sin1

sin

sinsin

Am

kxkxk

A

kxdxA

dxtkxAK

L

L

L

ω

µω

µω

µωω

=

−=

=

=

∫

∫

where m is the mass of the string. *69 •• Picture the Problem We can equate the expression for the velocity of a wave on a string and the expression for the velocity of a wave in terms of its frequency and wavelength to obtain an expression for the weight that must be suspended from the end of the string in order to produce a given standing wave pattern. By using the condition on the wavelength that must be satisfied at resonance, we can express the weight on the end of the string in terms of µ, f, L, and an integer n and then evaluate this expression for n = 1, 2, and 3 for the first three standing wave patterns. Express the velocity of a wave on the string in terms of the tension T in the string and its linear density µ:

µµmgTv ==

where mg is the weight of the object suspended from the end of the string.

Express the wave speed in terms of its wavelength λ and frequency f:

λfv =

Eliminate v to obtain:

µλ mgf =

Solve for mg: 22λµ fmg =

Express the condition on λ that corresponds to resonance:

... 3, 2, 1, ,2== n

nLλ

Substitute to obtain: ... 3, 2, 1, ,2 2

2 =⎟⎠⎞

⎜⎝⎛= n

nLfmg µ

or

... 3, 2, 1, ,42

22

== nn

Lfmg µ

Chapter 16

1278

Evaluate mg for n = 1: ( )( ) ( )( )

N425.0

1m2.0s80g/m415.04

2

221-

=

=mg

which corresponds, at sea level, to a mass of 43.3 g.

Evaluate mg for n = 2: ( )( ) ( )( )

N106.0

2m2.0s80g/m415.04

2

221-

=

=mg

which corresponds, at sea level, to a mass of 10.8 g.

Wave Packets 70 • Picture the Problem We can find the maximum duration of each pulse under the conditions given in the problem from the reciprocal of frequency of the pulses and the range of frequencies from the wave packet condition on ∆ω and ∆t. (a) The maximum duration of each pulse is its period:

s100.0s10s10

11 717 µ==== −

−fT

(b) Express the wave packet condition on ∆ω and ∆t:

1≈∆∆ tω or 12 ≈∆∆ tfπ

Solve for ∆f: ππ 22

1 Tt

f =∆

≈∆

Substitute numerical values and evaluate ∆f:

MHz59.12

s10 17

=≈∆−

πf

71 • Picture the Problem We can approximate the duration of the pulse from the product of the number of cycles in the interval and the period of each cycle and the wavelength from the number of complete wavelengths in ∆x. We can use its definition to find the wave number k from the wavelength λ.

(a) Relate the duration of the pulse to the number of cycles in the interval and the period of each cycle:

0fNNTt =≈∆


1279

(b) There are about N complete wavelengths in ∆x; hence:

Nx∆

≈λ

(c) Use its definition to express the wave number k:

xNk

∆==

πλπ 22

(d) defined.

not well is stops and starts pulse the wherehence, time;someat abruptly stoppingn rather thagradually out dies waveform thebecauseuncertain is N

(e) Using our result in part (c), express the uncertainty in k:

xx

Nk∆

=∆∆

=∆ππ 22

because ∆N = ±1. General Problems

72 • Picture the Problem We can use v = fλ and µFv = to relate the tension in the piano

wire to its fundamental frequency. Relate the tension in the wire to the speed of transverse waves on it:

mFLFv ==

µ

Express the speed of the transverse in terms of their wavelength and frequency:

λfv =

Equate these expressions and solve for F to obtain:

LmfF

22λ=

Relate λ for the fundamental mode of vibration to the length of the piano wire:

L2=λ


LmfF 24=

Substitute numerical values and evaluate F: ( )( ) ( )kN53.1

m8.0s63.261kg1074 213

=

×= −−F

Chapter 16

1280

73 • Picture the Problem We can use v = fnλn to express the resonance frequencies of the ear

canal in terms of their wavelengths and ... 5, 3, 1, ,4

== nnL nλto relate the length of the

ear canal to its resonance wavelengths. (a) Relate the resonance frequencies to the speed of sound and the wavelength of the compressional vibrations:

nn

vfλ

=

Express the condition for constructive interference in a pipe that is open at one end:

... 5, 3, 1, ,4

== nnL nλ

Solve for λn: nL

n4

=λ


( )( )kHz40.3

m102.54m/s340

4 2

n

nLvnfn

=×

== −

Evaluate f1, f2, and f3: kHz40.31 =f ,

kHz2.10kHz40.333 =×=f ,

and kHz0.17kHz40.355 =×=f

(b)

perceived.readily most be willHz 3400near sFrequencie

74 •

Picture the Problem We can use ... 5, 3, 1, ,4

== nnL nλto express the wavelengths of

the fundamental and next two harmonics in terms of the length of the rope and v = fnλn

and µFv = to relate the resonance frequencies to their wavelengths.

(a) Express the condition for constructive interference on a rope

... 5, 3, 1, ,4

== nnL nλ


1281

that is fixed at one end: Solve for λn: ( )

nnnL

nm16m444

===λ

Evaluate λn for n = 1, 3, and 5: m0.161 =λ

m33.53m16

3 ==λ

and

m20.35m16

5 ==λ

(b) Relate the resonance frequencies to the speed and wavelength of the transverse waves:

nn

vfλ

=

Express the speed of the transverse waves as a function of the tension in the rope:

mFLFv ==

µ

where m and L are the mass and length of the rope.


n

nnn m

FLf

λ

λλm/s100

kg0.16m4N40011

=

==

Evaluate fn for n = 1, 3, and 5: Hz25.6

m16m/s100

1 ==f

Hz8.18m33.5

m/s1003 ==f

and

Hz3.31m20.3

m/s1005 ==f

75 •• Picture the Problem The path difference at the point where the resultant wave an amplitude A is related to the phase shift between the interfering waves according to

πδλ 2=∆x . We can use this relationship to find the phase shift and the expression for

the amplitude resulting from the superposition of two waves of the same amplitude and frequency to find the phase shift.

Chapter 16

1282

Express the relation between the path difference and the phase shift at the point where the resultant wave has an amplitude A:

πδλ2

=∆x

Express the amplitude resulting from the superposition of two waves of the same amplitude and frequency:

δ21

0 cos2yA =

Solve for and evaluate δ: 3

22

cos22

cos2 1

0

1 πδ === −−

AA

yA


λπ

πλ 31

232

==∆x

76 •• Picture the Problem We can use v = fnλn to express the resonance frequencies of the

string in terms of their wavelengths and ... 3, 2, 1, ,2

== nnL nλto relate the length of the

string to the resonance wavelengths for a string fixed at both ends. Our strategy for part (b) will be the same … except that we’ll use the standing-wave condition

... 5, 3, 1, ,4

== nnL nλfor strings with one end free.

(a) Relate the frequencies of the harmonics to their wavelengths and the speed of transverse waves on the string:

nn

vfλ

=

Express the standing-wave condition for a string with both ends fixed:

... 3, 2, 1, ,2

== nnL nλ

Solve for λn: n

Ln

2=λ

Substitute to obtain: L

vnfn 2=

Express the speed of the transverse waves as a function of the tension in the string:

µFv =


1283


( )( )Hz657.0

kg/m0.0085N18

m3521

21

n

n

FL

nfn

=

=

=µ

Calculate the 1st four harmonics:

Hz657.01 =f

( ) Hz31.1Hz657.022 ==f

( ) Hz97.1Hz657.033 ==f

and ( ) Hz63.2Hz657.044 ==f

(b) Express the standing-wave condition for a string fixed at one end:

... 5, 3, 1, ,4

== nnL nλ

Solve for λn: n

Ln

4=λ

The resonance frequencies equation becomes:

( )( )Hz329.0

kg/m0.0085N18

m3541

41

n

n

FL

nfn

=

=

=µ

Calculate the 1st four harmonics:

Hz329.01 =f

( ) Hz987.0Hz329.033 ==f

( ) Hz65.1Hz329.055 ==f

and ( ) Hz30.2Hz329.077 ==f

77 •• Picture the Problem We’ll model the shaft as a pipe of length L with one end open. We can relate the frequencies of the harmonics to their wavelengths and the speed of sound using v = fnλn and the depth of the mine shaft to the resonance wavelengths using the

Chapter 16

1284

standing-wave condition for a pipe with one end open; ... 5, 3, 1, ,4

== nnL nλ.

Relate the frequencies of the harmonics to their wavelengths and the speed of sound:

nn

vfλ

=

Express the standing-wave condition for a pipe with one end open:

... 5, 3, 1, ,4

== nnL nλ

Solve for λn: n

Ln

4=λ


Lvnfn 4

=

For fn = 63.58 Hz:

Lvn

4Hz58.63 =

For fn+2 = 89.25 Hz: ( )

Lvn

42Hz25.89 +=

Divide either of these equations by the other and solve for n to obtain:

595.4 ≈=n

Substitute in the equation for fn = f5 = 63.58 Hz: L

vf45

5 =

Solve for and evaluate L: ( )

( ) m68.6s63.584

m/s340545

15

=== −fvL

78 •• Picture the Problem We can use the standing-wave condition for a string with one end free to find the wavelength of the 5th harmonic and the definitions of the wave number and angular frequency to calculate these quantitities. We can then substitute in the wave function for a wave in the nth harmonic to find the wave function for this standing wave. (a) Express the standing-wave condition for a string with one end free:

... 5, 3, 1, ,4

== nnL nλ


1285

Solve for and evaluate λ5: ( ) m00.45m54

54

5 ===Lλ

(b) Use its definition to calculate the wave number:

1

55 m

2m422 −===

ππλπk

(c) Using its definition, calculate the angular frequency:

( ) 1155 s800s40022 −− === πππω f

(d) Write the wave function for a standing wave in the nth harmonic:

( ) txkAtxy nnn ωcossin, =


( ) ( ) ( ) ( ) ( ) txtxkAtxy 11555 s800cosm

2sinm03.0cossin, −−

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛== ππω

79 •• Picture the Problem The coefficient of the factor containing the time dependence in the wave function is the maximum displacement of any point on the string. The time derivative of the wave function is the instantaneous speed of any point on the string and the coefficient of the factor containing the time dependence is the maximum speed of any point on the string. Differentiate the wave function with respect to t to find the speed of any point on the string:

[ ]( )( )

txtx

txt

vy

ππππππ

ππ

60sin4sin2.160sin4sin6002.0

60cos4sin02.0

−=−=∂∂

=

(a) Referring to the wave function, express the maximum displacement of the standing wave:

( ) ( ) ( )[ ]xxy 1max m4sinm02.0 −= π (1)

Evaluate equation (1) at x = 0.10 m: ( ) ( )( )( )[ ]cm90.1

m10.0m4sin

m02.0m10.01

max

=

×

=−π

y

Referring to the derivative of the wave function with respect to t, express the maximum speed of the

( ) ( ) ( )[ ]xxvy1

max, m4sinm/s2.1 −= ππ (2)

Chapter 16

1286

standing wave: Evaluate equation (2) at x = 0.10 m: ( ) ( )

( )( )[ ]m/s59.3

m10.0m4sin

m/s2.1m10.01

max,

=

×

=−π

πyv

(b) Evaluate equation (1) at x = 0.25 m:

( ) ( )( )( )[ ]

0

m25.0m4sin

m02.0m25.01

max

=

×

=−π

y

Evaluate equation (2) at x = 0.25 m: ( ) ( )

( )( )[ ]0

m25.0m4sin

m/s2.1m25.01

max,

=

×

=−π

πyv

(c) Evaluate equation (1) at x = 0.30 m:

( ) ( )( )( )[ ]cm18.1

m30.0m4sin

m02.0m30.01

max

=

×

=−π

y


( )( )[ ]m/s22.2

m30.0m4sin

m/s2.1m30.01

max,

=

×

=−π

πyv

(d) Evaluate equation (1) at x = 0.50 m:

( ) ( )( )( )[ ]

0

m50.0m4sin

m02.0m50.01

max

=

×

=−π

y


( )( )[ ]0

m50.0m4sin

m/s2.1m50.01

max,

=

×

=−π

πyv

80 •• Picture the Problem In part (a) we can use the standing-wave condition for a wire fixed at both ends and the fact that nodes are separated by one-half wavelength to find the harmonic number. In part (b) we can relate the resonance frequencies to their wavelengths and the speed of transverse waves and express the speed of the transverse


1287

waves in terms of the tension in the wire and its linear density. (a) Express the standing-wave condition for a wire fixed at both ends:

... 3, 2, 1, ,2

== nnL nλ

Solve for n: n

Lnλ2

=

Solve for and evaluate λ1: ( ) m5m5.2221 === Lλ

Relate the distance between nodes to the distance of the node closest to one end and solve for λn:

m5.021 =nλ

and m1=nλ

Substitute and evaluate n: ( ) 5

m1m5.22

==n

(b) Express the resonance frequencies in terms of the their wavelengths and the speed of transverse waves on the wire:

1λλvnvf

nn ==

Relate the speed of transverse waves on the wire to the tension in the wire:

µFv =

Substitute and simplify to obtain: ( )( )

( )Hz48.5kg0.1

m2.5N30m511

1

n

nmFLnfn

=

==λ

Evaluate fn for n = 1, 2, and 3: Hz48.51 =f

( ) Hz0.11Hz48.522 ==f

and ( ) Hz4.16Hz48.533 ==f

*81 •• Picture the Problem We can use λfv = to relate the speed of sound in the gas to the

distance between the piles of powder in the glass tube.

Chapter 16

1288

(a)nodes. at the collectsit so and

,stationary ispowder thenode aat about; moved ispowder theantinode,nt displaceme aAt tube.in the upset are wavesstanding resonance,At

(b) Relate the speed of sound to its frequency and wavelength:

λfv =

Letting D = distance between nodes, relate the distance between the nodes to the wavelength of the sound:

D2=λ


fDv 2=

(c) If we let the length L of the tube be 1.2 m and assume that vair = 344 m/s (the speed of sound in air at 20°C), then the 10th harmonic corresponds to D = 25.3 cm and a driving frequency of:

( ) Hz680 m253.02

m/s3442

airair ===

Dvf

(d)

helium.or air either in sound of speed theoft measuremen for the work well wouldcm) (218 cm 25.3 and

cm 8.60 of multiplecommon least theislength whose tubea end,driven at the effects end neglecting Hence, cm. 8.60 be air wouldin harmonic

10 for the and cm 25.3 wouldheliumin harmonic 10 for the then

)C20at heliumin sound of speed (the m/s 1008 and kHz 2 Ifthth

He

DD

vf ,°==

82 •• Picture the Problem We can use µFv = to express F as a function of v and

λfv = to relate v to the frequency and wavelength of the string’s fundamental mode.

Because, for a string fixed at both ends, fn = nf1, we can extend our result in part (a) to part (b). (a) Relate the speed of the transverse waves on the string to the tension in it:

µFv =

Solve for F: 2vF µ= (1)


1289

Relate the speed of the transverse waves on the string to their frequency and wavelength:

11λfv =

Express the wavelength of the fundamental mode to the length of the string:

L21 =λ

Substitute to obtain: fLv 2=


µ224 LfF = (2)


( ) ( ) ( )N720

kg/m108m5.2s604 3221

=

×= −−F

(b) For the nth harmonic, equation (2) becomes:

( )N7202221

222 nLfnLfF nn === µµ

Evaluate this expression for n = 2, 3, and 4:

( ) kN88.2N72042 ==F

( ) kN48.6N72093 ==F

and ( ) kN5.11N720164 ==F

83 •• Picture the Problem We can use the conditions 1ff =∆ and 1nffn = , where n is an

integer, which must be satisfied if the pipe is open at both ends to decide whether the pipe is closed at one end or open at both ends. Once we have decided this question, we can use the condition relating ∆f and the fundamental frequency to determine the latter. In part (c) we can use the standing-wave condition for the appropriate pipe to relate its length to its resonance wavelengths. (a) Express the conditions on the frequencies for a pipe that is open at both ends:

1ff =∆

and 1nffn =

Evaluate ∆f = f1: Hz524Hz1310Hz8341 =−=∆f

Using the 2nd condition, find n:

5.2Hz524Hz1310

1

===ffn n

Chapter 16

1290

end. oneat closed is pipe The

(b) Express the condition on the frequencies for a pipe that is open at both ends:

12 ff =∆

Solve for and evaluate f1:

( ) Hz262Hz52421

21

1 ==∆= ff

(c) Using the standing-wave condition for a pipe open at one end, relate the length of the pipe to its resonance wavelengths:

... 5, 3, 1, ,4

== nnL nλ

For n = 1 we have:

11 f

v=λ and

1

1

44 fvL ==

λ

Substitute numerical values and evaluate L: ( ) cm32.4

s2624m/s340

1 == −L

84 •• Picture the Problem We can relate the speed of sound in air to the frequency of the violin string and the wavelength of the sound in the open tube that is closed at one end by water. The wavelength of the sound, in turn, is a function of the length of the air column and so we can derive an expression for the speed of sound as a function of the frequency of the transverse waves on the violin string and the length of the air column above the water. Knowing that the violin string is vibrating in its fundamental mode, we can express this frequency in terms of the tension in the string and its linear density. Express the speed of sound in the tube in terms of its fundamental frequency and wavelength:

11s λfv =

Using the standing-wave condition for a tube open at one end, relate the speed of sound to the length of the air column in the tube:

... 5, 3, 1, ,4columnair == nnL nλ

Solve for λ1: columnair1 4L=λ

Substitute to obtain: columnair1s 4 Lfv = (1)


1291

Express the frequency of the transverse waves on the violin string in terms of their wavelength and the speed with which they propagate on the string:

string11 2L

vvf ==λ

Relate the speed of the transverse waves on the string to the tension in it:

mFLFv string==

µ


string

string

string1 42

1mL

Fm

FLL

f ==


stringcolumnair

stringcolumnairs

2

44

mLFL

mLFLv

=

=

Substitute numerical values and evaluate vs:

( ) ( )( )( )

m/s338

m0.5kg10N440m0.182 3s

=

= −v

56). Problem (see effects end neglectsit because accurate not very is method The

85 •• Picture the Problem We know that the superimposed traveling waves have the same wave number and angular frequency as the standing-wave function, have equal amplitudes that are half that of the standing-wave function, and travel in opposite directions. From inspection of the standing-wave function we note that

121 m−= πk and 1s40 −= πω . We can express the velocity of a segment of the rope by

differentiating the standing-wave function with respect to time and the acceleration by differentiating the velocity function with respect to time. (a) Write the wave function for the wave traveling in the positive x direction:

( ) ( ) ( ) ⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛= −− txtxy 11

1 s40m2

sinm01.0, ππ

Chapter 16

1292

Write the wave function for the wave traveling in the negative x direction:

( ) ( ) ( ) ⎥⎦

⎤⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛= −− txtxy 11

2 s40m2

sinm01.0, ππ

(b) Express the distance d between nodes in terms of the wavelength of the standing wave:

λ21=d

Use the wave number to find the wavelength:

λππ 2m 1

21 == −k

and m4=λ

Substitute and evaluate d: ( ) m2.00m42

1 ==d

(c) Differentiate the given wave function with respect to t to express the velocity of any segment of the rope:

( ) ( ) ( )

( ) ( )tx

txt

txvy

11

11

s40sinm2

sinm/s8.0

s40cosm2

sinm02.0,

−−

−−

⎟⎠⎞

⎜⎝⎛−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

∂∂

=

πππ

ππ

Evaluate ( )tvy ,m1 :

( ) ( ) ( ) ( )( ) ( )

( ) ( )tt

ttvy

1

1

11

s40sinm/s51.2

s40sinm/s8.0

s40sinm1m2

sinm/s8.0,m1

−

−

−−

−=

−=

⎟⎠⎞

⎜⎝⎛−=

π

ππ

πππ

(d) Differentiate ( )txvy , with respect to time to obtain ( )txay , :

( ) ( ) ( )

( ) ( )tx

txt

txay

1122

11

s40cosm2

sinm/s32

s40sinm2

sinm/s8.0,

−−

−−

⎟⎠⎞

⎜⎝⎛−=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

∂∂

=

πππ

πππ


1293

Evaluate ( )tay ,m1 :

( ) ( ) ( ) ( )( ) ( )

( ) ( )tt

ttay

12

122

1122

s40cosm/s316

s40cosm/s32

s40cosm1m2

sinm/s32,m1

−

−

−−

−=

−=

⎟⎠⎞

⎜⎝⎛−=

π

ππ

πππ

86 •• Picture the Problem We can use the definition of intensity to find the intensity of each speaker, the dependence of intensity on the square of the amplitude of the wave disturbance to express the amplitudes of the waves, and the dependence of the intensity on whether the speakers are coherent and their phase difference to find the intensity at the given point. (a) Express the intensity as a function of the distance of a point from the source:

24 rPIπ

=

Evaluate I1: ( )

221 W/m9.19

m24mW1 µ

π==I

Evaluate I2:

( )2

22 W/m84.8m34

mW1 µπ

==I

(b) Using v = fλ, find the wavelength of the sound:

m5.0s680m/s340

1 === −fvλ

Express the path difference in terms of λ:

λ2=∆x and so there is constructive interference at point P.

Express the intensity at point P due to the sound from source 1:

211 constant AI ×=

or

11 ICA =

where C is a constant.

Express the intensity at point P due the sound from source 2:

222 constant AI ×=

or

22 ICA =

Chapter 16

1294

Express the square of the resultant amplitude at point P:

( ) ICIICA 22

2122 =+=

Solve for and evaluate I: ( )( )

2

222

2

21

W/m3.55

W/m84.8W/m9.19

µ

µµ

=

+=

+= III

(c) If they are driven coherently but are 180° out of phase we will have destructive interference at point P and the intensity is given by:

( )( )

2

222

2

21

W/m21.2

W/m84.8W/m9.19

µ

µµ

=

−=

−= III

(d) Because the sources are incoherent, the intensities add arithmetically: 2

2221

W/m7.28

W/m84.8W/m9.19

µ

µµ

=

+=

+= III

87 •• Picture the Problem In Chapter 14, Section 14.1, it was shown that a harmonic function could be represented by a vector rotating at the angular frequency ω. The simplest way to do this problem is to use that representation. The vectors, of equal magnitude, are shown in the diagram. We can find the resultant wave function by finding the magnitude and direction of the resultant vector.

From the diagram it is evident that: ∑ = 0yv

Find the sum of the x components of the vectors:

AAAAvx 260cos60cos =+°+°=∑

Relate the magnitude of the resultant vector to the sum of its x and y components:

( ) ( )( ) ( ) AA

vvv yx

202 22

22

=+=

+= ∑∑


1295

Find the direction of the resultant vector: 0

20tantan 11 =⎟

⎠⎞

⎜⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛= −−

∑∑

Avv

x

yθ

Express the resultant wave: ( ) ( )( )tkx

tkxAtxy

ω

ω

−=

−=

sin1.0

sin2,res

88 •• Picture the Problem The diagram shows a two dimensional plane wave propagating at an angle θ with respect to the x axis. At a given point in time, the surface of constant phase for the wave is the line defined by kx x + kyy = φ , or ( ) φ+−= xkky yx .

The wave itself moves in a direction perpendicular to the wavefront, i.e., in a direction specified by a line with slope ky/kx. Choose two points (x, y) and (x + ∆x, y + ∆y) that have a separation of 1 wavelength along such a line.

Express the phase difference φ between the two points that have a separation of 1 wavelength along the line ( ) φ+−= xkky yx in terms

of the spatial separation ∆r of the points:

λπφ 2

=∆r

or r∆=λπφ 2

where ( ) ( )22 yxr ∆+∆=∆

Substitute φ = 2π to obtain: ( ) ( )2222 yx ∆+∆=λππ

or

( ) ( )22 yx ∆+∆=λ (1)

Express φ in terms of kx, ky, ∆x and ∆y: ykxkrk yx ∆+∆=∆=φ

or, because φ = 2π, π2=∆+∆ ykxk yx

Because xkk

yx

y ∆=∆ : π22

=∆+∆ xkk

xkx

yx

or

222

yx

x

kkkx

+=∆

π

Chapter 16

1296

Similarly: 22

2

yx

y

kk

ky

+=∆

π


22

2

22

2

22

2

22

yx

yx

y

yx

x

kk

kkk

kkk

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛

++⎟

⎟⎠

⎞⎜⎜⎝

⎛

+=

π

ππλ

Relate the wave velocity v to its angular frequency ω and wave number k:

πλωω

2==

kv

Substitute for λ to obtain: 2222

22

yxyx kkkkv

+=

+=

ωππ

ω

Express the angle between the wave velocity and the x axis:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

+

+=

∆∆

=

−

−−

x

y

yx

x

yx

y

kk

kkkkk

k

xy

1

22

2211

tan

2

2

tantan π

π

θ

*89 •• Picture the Problem We can express the fundamental frequency of the organ pipe as a function of the air temperature and differentiate this expression with respect to the temperature to express the rate at which the frequency changes with respect to temperature. For changes in temperature that are small compared to the temperature, we can approximate the differential changes in frequency and temperature with finite changes to complete the derivation of ∆f/f = ½∆T/T. In part (b) we’ll use this relationship and the data for the frequency at 20°C to find the frequency of the fundamental at 30°C. (a) Express the fundamental frequency of an organ pipe in terms of its wavelength and the speed of sound:

λvf =

Relate the speed of sound in air to the absolute temperature: TC

MRTv ==

γ


1297

where

constant==MRC γ

Defining a new constant C′, substitute to obtain:

TCTCf '==λ

because λ is constant for the fundamental frequency we ignore any change in the length of the pipe.

Differentiate this expression with respect to T:

TfTC

dTdf

2'

21 21 == −

Separate the variables to obtain: T

dTf

df21

=

For ∆T << T, we can approximate df by ∆f and dT by ∆T to obtain:

TT

ff ∆

=∆

21

(b) Express the fundamental frequency at 30°C in terms of its frequency at 20°C:

fff ∆+= 2030

Solve our result in (a) for ∆f: TTff ∆

=∆ 21

Substitute numerical values and evaluate ∆f:

( )

Hz203

K293K10Hz200Hz200 2

130

=

+=f

90 •• Picture the Problem We’ll use a spreadsheet program to graph the wave functions individually and their sum as functions of x at t = 0 and at t = 1 s. In (c) and (d) we can add the wave functions algebraically to find the result wave function at t = 0 and at t = 1 s. (a) and (d) A spreadsheet program to calculate values for y1(x,t) and y2(x,t) between and plot their graphs is shown below. The constants and cell formulas used are shown in the table.

Cell Content/Formula Algebraic Form A5 −5.0 x A6 A5+0.1 xx ∆+

Chapter 16

1298

B5 0.05/(2+(A5−2*$B$1)^2) ( )0,1 xy C5 −0.05/(2+(A5+2*$B$1)^2) ( )0,2 xy D5 0.05/(2+(A5−2*$B$1)^2)

−0.05/(2+(A5+2*$B$1)^2)( ) ( )0,0, 21 xyxy +

E5 0.05/(2+(A5−2*$B$2)^2) −0.05/(2+(A5+2*$B$2)^2)

( ) ( )1,1, 21 xyxy +

A B C D E 1 t= 0 2 t= 1 s 3 4 x y1(x,0) y2(x,0) y1(x,0)+y2(x,0) y1(x,1)+y2(x,1) 5 −5.0 0.001 −0.001 0.000 −0.001 6 −4.9 0.001 −0.001 0.000 −0.002 7 −4.8 0.001 −0.001 0.000 −0.002 8 −4.7 0.001 −0.001 0.000 −0.002 9 −4.6 0.001 −0.001 0.000 −0.002

10 −4.5 0.001 −0.001 0.000 −0.002

110 5.5 0.001 −0.001 0.000 0.001 111 5.6 0.001 −0.001 0.000 0.001 112 5.7 0.001 −0.001 0.000 0.001 113 5.8 0.001 −0.001 0.000 0.001

The four curves on the graph are identified in the legend. y1 is traveling from left to right and y2 from right to left. As time increases, y1 is farther to the right and y2 is farther to the left.

-0.010

-0.008

-0.006

-0.004

-0.002

0.000

0.002

0.004

0.006

0.008

0.010

-5 -3 -1 1 3 5

y1(x,0)

y2(x,0)

y1(x,0)+y2(x,0)

y1(x,1)+y2(x,1)


1299

(b) Express the resultant wave function at t = 0:

( ) ( ) 0m2

m02.0m2

m02.00,0, 22

3

22

3

21 =+

−+

+=+

xxxyxy

(c) Express the resultant wave function at t = 1 s:

( ) ( )( ) ( )22

3

22

3

21 s2m2m02.0

s2m2m02.0s1,s1,

++−

+−+

=+xx

xyxy

91 •• Picture the Problem We can relate the frequency of the standing waves in the open-ended tube to its length and the speed of sound in air. (a) What you hear is the fundamental mode of the tube and its overtones. A more physical explanation is that the echo of the finger snap moves back and forth along the tube with a characteristic time of 2L/c, leading to a series of clicks from each echo. Because the clicks happen with a frequency of c/2L, the ear interprets this as a musical note of that frequency. (b) Express the frequency of the sound in terms of the length of the tube:

Lvf

2=

Solve for L: f

vL2

=

Substitute numerical values and evaluate L: ( ) cm6.38

s4402m/s340

1 == −L

92 •• Picture the Problem To find the total kinetic energy of the nth mode of vibration, we’ll need to differentiate ( ) txkAtxy nnnn ωcossin, = with respect to time, substitute in the

expression for ∆K, and then integrate over the length of the string. (a) Write the wave function for a standing wave on a string fixed at both ends:

( ) txkAtxy nnnn ωcossin, =

where n

nkλπ2

= .

Using the standing-wave condition for a string with both ends fixed, relate the length of the string to the

... 3, 2, 1, ,2

== nnL nλ

Chapter 16

1300

wavelength of the nth harmonic: Solve for λn:

nL

n2

=λ

Substitute in the expression for kn to obtain: L

nknπ

=

Differentiate this expression with respect to t:

[ ]txkA

txkAtt

y

nnnn

nnn

ωω

ω

sinsin

cossin

−=∂∂

=∂∂

Substitute in the given expression and simplify to obtain:

( )xtxkA

xtxkAK

nnnn

nnnn

∆=

∆−=∆

ωµω

ωωµ2222

21

221

sinsin

sinsin

Integrate this expression over the length of the string to find its total kinetic energy: tAm

dxxL

ntAK

nnn

L

nnn

ωω

πωµω

22241

0

222221

sin

sinsin

=

⎟⎠⎞

⎜⎝⎛= ∫

(b) Express the condition that K = Kmax:

1sin2 =tnω (1)


2241

max nn AmK ω=

(c) From equation (1), for K = Kmax:

1sin2 =tnω or 2πω =tn

Evaluate the wave function in (a)

when 2πω =tn :

02

cossin2

, ==⎟⎟⎠

⎞⎜⎜⎝

⎛ πωπ xkAxy nn

nn

(d) Using the result from part (b), express the maximum kinetic energy:

2241

max nn AmK ω=

Relate ωn to ω1: 1ωω nn =

Substitute to obtain: ( )22

1412

max nAmnK ω=

or, because m and ω1 are constants,


1301

22max nAnK ∝

Remarks: Our result in part (b) is exactly the same result obtained in Problem 68 with ωn and An replacing ω and A. 93 ••

Picture the Problem We can use ... 3, 2, 1, ,2

== nLvnfn to relate the resonant

frequencies to the length of the string and the speed of transverse waves on the string and µFv = to express the speed of the transverse waves on the string in terms of the

tension in the string. Differentiating of the resulting expression with respect to F will lead

toF

dFf

df

n

n

21

= . For changes in f that are small compared to f, we can use a differential

approximation to obtainFF

ff

n

n ∆=

∆21

.

(a) Using the standing-wave condition for a string fixed at both ends, relate the resonant frequencies to the length of the string and the speed of transverse waves on the string:

... 3, 2, 1, ,2

== nLvnfn

Express the speed of transverse waves on the string in terms of the tension in the string:

µFv =

Substitute to obtain: FCF

Lnfn ==

µ2

because n, L, and µ are constants.

Differentiate fn with respect to F to obtain: F

fF

CdFdf nn

211

2==


FdF

fdf

n

n

21

=

harmonics. allfor validis expression this,derivation itson placed wereconditions no Because

Chapter 16

1302

(b) Because ∆f << f, one can approximate the differential quantitities in our result for part (a) to obtain:

FF

ff

n

n ∆=

∆21

Solve for ∆F/F: n

n

ff

FF ∆

=∆ 2

Substitute numerical values and evaluate ∆F/F:

%54.1Hz260

Hz22 =⎟⎟⎠

⎞⎜⎜⎝

⎛=

∆FF

94 •• Picture the Problem Let the sources be denoted by the numerals 1 and 2. The phase difference between the two waves at point P is the sum of the phase difference due to the sources δ0 and the phase difference due to the path difference δ. (a) Write the wave function due to source 1:

( ) ( )tkxAtxf ω−= 101 cos,

Write the wave function due to source 2:

( ) ( )( )s102 cos, δω +−∆+= txxkAtxf

(b) Express the sum of the two wave functions:

( ) ( ) ( ) ( ) ( )( )( )[ ( )( )]s110

s101021

coscoscoscos,,,

δωωδωω

+−∆+−=+−∆++−=+=

txxktkxAtxxkAtkxAtxftxftxf

Use ⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +

=+2

cos2

cos2coscos βαβαβα to obtain:

( ) ⎥⎦

⎤⎟⎟⎠

⎞⎜⎜⎝

⎛+−⎟

⎠⎞

⎜⎝⎛ ∆

+⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +

∆=

22cos

22cos2, ss

0δωδ txxkxkAtxf

Express the phase difference δ in terms of the path difference ∆x and the wave number k:

kx

==∆ λ

πδ 2 or δ=∆xk


1303


( ) ⎥⎦

⎤⎟⎟⎠

⎞⎜⎜⎝

⎛+−⎟

⎠⎞

⎜⎝⎛ ∆

+⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +

=22

cos2

cos2, ss0

δωδδ txxkAtxf

The amplitude of the resultant wave function is the coefficient of the time-dependent factor:

( )s21

0 cos2 δδ += AA

(c) Express the intensity at an arbitrary point P: ( )[ ]

( )[ ]s

s

P

AC

AC

ACI

δδ

δδ

+=

+=

=

2122

0

221

0

2

cos4'

cos2'

'

Evaluate I for δ = 0 and δs = Ct: ( )[ ]CtACI 2

1220 cos4'=

Because the average value of θ2cos over a complete period is ½:

020ave 22 IAI =∝

and ( )CtII 2

120 cos4∝

(d) Evaluate I for λ2

1=∆x and

δs = Ct:

πδλ =⇒=∆ 21x

∴ ( )[ ]CtACI += π2122

0 cos4'

and at t = 0, 0=I . i.e., the waves interfere destructively.

A spreadsheet program to calculate the intensity at point P as a function of time for a zero path difference and a path difference of λ is shown below. The constants and cell formulas used are shown in the table.

Cell Content/Formula Algebraic Form B1 1 C B7 B6+0.1 tt ∆+ C6 COS($B$6*B6/2)^2 ( )Ct2

12cos D6 COS($B$6*B6/2-PI()/2)^2 ( )Ct+π2

12cos

A B C D 1 C= 1 s−1 2 3 4 t I I 5 (s) (W/m^2) (W/m^2) 6 0.00 1.000 0.000

Chapter 16

1304

7 0.10 0.998 0.002 8 0.20 0.990 0.010 9 0.30 0.978 0.022

103 9.70 0.019 0.981 104 9.80 0.035 0.965 105 9.90 0.055 0.945 106 10.00 0.080 0.920

The solid curve is the graph of ( )Ct2

12cos and the dashed curve is the graph of

( )Ct+π212cos .

0.0

0.2

0.4

0.6

0.8

1.0

0 2 4 6 8 10t (s)

I (W

/m^2

)

path diff = 0

path diff =lambda/2

95 ••• Picture the Problem We can differentiate the sum of the two wave functions to find the velocity of a segment dx of the string. We can find the kinetic energy of this segment from dxvdmvdK yy

2212

21 µ== and integrate this expression from 0 to L to find the total

kinetic energy of the resultant wave. (a) Express the resultant wave function:

( ) ( ) ( ) xktAxktAtxytxytxy 22211121r sincossincos,,, ωω +=+=

Differentiate this expression with respect to t to find vy:

( ) [ ]

xktAxktA

xktAxktAt

txvy

22221111

222111

sinsinsinsin

sincossincos,

ωωωω

ωω

−−=

+∂∂

=


1305

(b) Express the kinetic energy of a segment of the string of length dx and mass dm:

( )[

]dxxktA

xktxktAAxktA

dxxktAxktAdxvdmvdK yy

22

222

222

2211212112

122

1212

1

2222211112

12212

21

sinsin

sinsinsinsin2sinsin

sinsinsinsin

ωω

ωωωωωωµ

ωωωωµµ

+

+=

+===

(c) Integrate dK from 0 to L to obtain:

( )( ) ( )( )( )( )

tAmtAm

LtA

ttAALtA

xdxL

ntA

xdxL

nxL

nttAA

xdxL

ntA

xdxktA

xdxktxktAA

xdxktAK

nn

L

L

L

L

L

L

222

2224

11

221

214

1

21

222

2222

1

21212121

122

1212

1

02

22

222

222

1

021212121

01

21

221

212

1

02

22

222

222

1

0221121212

1

01

21

221

212

1

sinsin

sin

0sinsinsin

sinsin

sinsinsinsin

sinsin

sinsin

sinsinsinsin2

sinsin

21

ωωωω

ωµω

ωωωµωωµω

πωµω

ππωωωµω

πωµω

ωωµ

ωωωωµ

ωωµ

+=

+

+=

+

+

=

+

+

=

≠

∫

∫

∫

∫

∫

∫

Note that, from Problem 92: 212

222

224

11

221

214

1 sinsin KKtAmtAm +=+ ωωωω

96 ••• Picture the Problem We can use the relationship 22

41

max AmK ω= from Problem 92 to

express the maximum kinetic energy of the wire and v = fλ and µFv = to find an

expression for ω. In part (d) we’ll use 2

21 ⎟

⎠⎞

⎜⎝⎛

∂∂

≈∆∆

xyF

xU

from Problem 15-120 to

determine where the potential energy per unit length has its maximum value. (a) From Problem 92 we have: 22

41

max AmK ω= (1)

Chapter 16

1306

Express ω1 in terms of f1: 11 2 fπω =

Relate f1 to the speed of transverse waves on the wire and the wavelength of the fundamental mode:

Lvvf

211 ==

λ

where L is the length of the wire.

Express the speed of the transverse waves on the wire in terms of the tension in the wire:

mFLFv ==

µ


mLF

mFL

Lf

421

1 ==

Substitute for ω1 and f1 in equation (1) to obtain:

22

2

2

41

max 442 A

LFA

mLFmK ππ =⎥

⎦

⎤⎢⎣

⎡=

Substitute numerical values and evaluate Kmax:

( )( ) ( )

mJ7.19

m102m24

N40 222

max

=

×= −πK

(b) Express the wave function for a standing wave in its first harmonic:

( ) txkAtxy 1111 cossin, ω= (2)

At the instant the transverse displacement is given by (0.02 m) sin (πx/2):

01cos 11 =⇒= tt ωω

and 0=K

(c) dK is a maximum where the displacement of the wire is greatest; i.e., at its midpoint:

( ) m00.1m221

21 === Lx

(d) From Problem 15-120: 2

21 ⎟

⎠⎞

⎜⎝⎛

∂∂

≈∆∆

xyF

xU

Express the condition on xy ∂∂ that

maximizes ∆U/∆x: max⎟⎠⎞

⎜⎝⎛

∂∂

=∂∂

xy

xy


1307

Differentiate ( ) txkAtxy 1111 cossin, ω= with

respect to x and set the derivative equal to zero for extrema:

( )

0coscos

cossin

1111

1111

==

∂∂

=∂∂

txkAk

txkAxx

y

ω

ω

or 0cos 1 =xk

Solve for k1x and then x:

21π

=xk

and

( ) ( )

( ) m00.1m2

2222

21

41

41

1

==

==== Lk

x λπ

πλπ

i.e., the potential energy per unit length is a maximum at the midpoint of the wire.

Remarks: In part (d) we’ve shown that ∆U/∆x has an extreme value at x = 1 m. To show that ∆U/∆x is a maximum at this location, you need to examine the sign of the 2nd derivative of y1(x,t) at this point. 97 ••• (a) A spreadsheet program to evaluate f(x) is shown below. Typical cell formulas used are shown in the table.

Cell Content/Formula Algebraic Form A6 A5+0.1 xx ∆+ B4 2*B3+1 12 +n B5 (−1)^B$3*COS(B$4*$A5)

/B$4*4/PI()

( )( )∑∞

= ++−

0 1212cos)1(4

n

n

nxn

π

C5 B5+(−1)^C$3*COS(C$4*$A5) /C$4*4/PI()

( )( )∑∞

= ++−

0 1212cos)1(4

n

n

nxn

π

A B C D K L 1 2 3 0 1 2 9 10 4 1 3 5 19 21 5 0.0 1.2732 0.8488 1.1035 0.9682 1.0289 6 0.1 1.2669 0.8614 1.0849 1.0134 0.9828 7 0.2 1.2479 0.8976 1.0352 1.0209 0.9912 8 0.3 1.2164 0.9526 0.9706 0.9680 1.0286 9 0.4 1.1727 1.0189 0.9130 1.0057 0.9742

10 0.5 1.1174 1.0874 0.8833 1.0298 1.0010

Chapter 16

1308

130 12.5 1.2704 0.8544 1.0952 0.9924 1.0031 131 12.6 1.2725 0.8503 1.1013 0.9752 1.0213 132 12.7 1.2619 0.8711 1.0710 1.0287 0.9714 133 12.8 1.2386 0.9143 1.0141 1.0009 1.0126 134 12.9 1.2030 0.9740 0.9493 0.9691 1.0146 135 13.0 1.1554 1.0422 0.8990 1.0261 0.9685

The solid curve is plotted from the data in columns A and B and is the graph of f(x) for 1 term. The dashed curve is plotted from the data in columns A and F and is the graph of f(x) for 5 terms. The dotted curve is plotted from the data in columns A and K and is the graph of f(x) for 10 terms.

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0 2 4 6 8 10 12 14

n = 1n = 5n = 10

(b) Evaluate f(2π) to obtain: ( )

( )

1

...71

51

3114

...5

25cos3

23cos12cos4)2(

=

⎟⎠⎞+−+⎜

⎝⎛ −=

⎟⎠⎞−+

⎜⎝⎛ −=

π

π

πππ

πf

which is equivalent to the Liebnitz formula.

98 ••• (a) A spreadsheet program to evaluate f(x) is shown below. Typical cell formulas used are shown in the table.

Cell Content/Formula Algebraic Form A6 A5+0.1 xx ∆+ B4 2*B3+1 12 +n B5 (−1)^$B$3*sin($B$4*A5)/

($B$4)^2*4/PI() ( ) ( )

( )∑ ++−

n

n

nxn

21212sin14

π


1309

C5 B5+((−1)^$C$3*sin($C$4*A5)/ ($C$4)^2*4/PI()

( ) ( )( )∑ +

+−

n

n

nxn

21212sin14

π

A B C D K L 1 2 3 0 1 2 9 10 4 1 3 5 19 21 5 0.0 0.0000 0.0000 0.0000 0.0000 0.0000 6 0.1 0.1271 0.0853 0.1097 0.0986 0.1011 7 0.2 0.2530 0.1731 0.2159 0.2012 0.1987 8 0.3 0.3763 0.2654 0.3163 0.3004 0.3005 9 0.4 0.4958 0.3640 0.4103 0.3983 0.4008 10 0.5 0.6104 0.4693 0.4998 0.5011 0.4985

72 6.7 0.5155 0.3812 0.4256 0.4153 0.4171 73 6.8 0.6291 0.4877 0.5146 0.5183 0.5154 74 6.9 0.7365 0.6005 0.6034 0.6171 0.6182 75 7.0 0.8365 0.7181 0.6963 0.7148 0.7166 76 7.1 0.9282 0.8380 0.7968 0.8183 0.8155

Graphs of f(x) for 1, 5, and 10 terms are shown below. Note that there is little difference between the graphs for 5 terms and 10 terms of this triangular wave function.

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

0 1 2 3 4 5 6 7

n = 1n = 5n = 10

99 ••• Picture the Problem From the diagram above, the nth echo will reflect n − 1 times going out, and the same number of times going back. If we "unfold" the ray into a straight line, we get the representation shown below. Using this figure we can express the distance dn traveled by the nth echo and then use this result to express the time delay between the nth and n + 1th echoes. The reciprocal of this time delay is the frequency corresponding to the nth echo.

Chapter 16

1310

(a) Apply the Pythagorean theorem to the right triangle whose base is L, whose height is 2(n − 1), and whose hypotenuse is dn to obtain:

222)1(42 Lrndn +−=

Express the time delay between the nth and n + 1th echoes: v

dt nn =∆


( )[ ] ⎟⎠⎞+−−

⎜⎝⎛ +=∆

222

222

12

22

Lrn

Lrnv

tn

A spreadsheet program to calculate ∆tn as a function of n is shown below. The constants and cell formulas used are shown in the table.

Cell Content/Formula Algebraic Form B1 90 L B2 1 r B3 340 c B8 B7+1 n + 1 C7 2/$B$3*((2*(B7−1)

*$B$2)^2+$B$1^2)^0.5 nt∆

A B C D 1 L= 90 m 2 r= 1 m 3 c= 340 m/s 4 5 6 n t(n) delta t(n) 7 1 0.5294 0.0001 8 2 0.5295 0.0004 9 3 0.5299 0.0007

10 4 0.5306 0.0009 11 5 0.5315 0.0012


1311

202 196 2.3544 0.0115 203 197 2.3659 0.0115 204 198 2.3773 0.0115 205 199 2.3888 0.0115 206 200 2.4003 0.0115

The graph of ∆tn as a function of n shown below was plotted using the data from columns B and D.

0.000

0.002

0.004

0.006

0.008

0.010

0.012

0 50 100 150 200

n

Del

ta-t

(n)

(c) decreases.istler culvert wh theoffrequency thetime,

over increases because so ,1/ is any timeat heardfrequency The nn tt ∆∆

The highest frequency corresponds to n = 1 and is given by:

1highest

1t

f∆

=

Substitute for ∆t1 to obtain:

( ) )⎜⎝⎛ −+

=∆

=2222

1highest

22

1

LLr

vt

f

Substitute numerical values and evaluate fhighest: ( ) ( ) )

kHz65.7

m90m90m142

m/s34022highest

=

⎜⎝⎛ −+

=f

The lowest frequency end can be found by examining the limit of ∆tn as n → ∞:

( )( )

( )( )( )

( )vrnn

vr

nLrn

nLrn

vt nnn

42222

1212

222limlim 2

22

2

22

=+−=

⎥⎥⎦

⎤

⎟⎟⎠

⎞

−+−−

⎢⎢⎣

⎡

⎜⎜⎝

⎛+=∆ →∞→∞

Chapter 16

1312

Express flowest in terms of ∆t∞: r

vt

f4

1lowest =

∆=

∞

Substitute numerical values and evaluate flowest: ( ) Hz0.85

m14m/s340

lowest ==f

1313

Chapter 17 Temperature and the Kinetic Theory of Gases Conceptual Problems

*1 • (a) False. If two objects are in thermal equilibrium with a third, then they are in thermal equilibrium with each other. (b) False. The Fahrenheit and Celsius temperature scales differ in the number of intervals between the ice-point temperature and the steam-point temperature. (c) True. (d) False. The result one obtains for the temperature of a given system is thermometer-dependent.

2 • Determine the Concept Put each in thermal equilibrium with a third body; e.g., a thermometer. If each body is in thermal equilibrium with the third, then they are in thermal equilibrium with each other. 3 • Picture the Problem We can decide which room was colder by converting 20°F to the equivalent Celsius temperature.

Using the Fahrenheit-Celsius conversion, convert 20°F to the equivalent Celsius temperature:

( ) ( )C67.6

322032 95

F95

C

°−=

°−°=°−= tt

so colder. wasroom sMert'

4 ••

Picture the Problem We can apply the ideal-gas law to the two vessels to decide which of these statements is correct.

Apply the ideal-gas law to the particles in vessel 1:

1111 kTNVP =

Apply the ideal-gas law to the particles in vessel 2:

2222 kTNVP =

Divide the equation for vessel 1 by the equation for vessel 2: 22

11

22

11

kTNkTN

VPVP

=

Chapter 17

1314

Because the vessels are identical and are at the same temperature and pressure:

2

11NN

= and 21 NN =

( ) correct. is a

5 ••

Determine the Concept From the ideal-gas law, we have .VTnRP = In the process

depicted, both the temperature and the volume increase, but the temperature increases faster than does the volume. Hence, the pressure increases. *6 •• Determine the Concept From the ideal-gas law, we have .PTnRV = In the process depicted, both the temperature and the pressure increase, but the pressure increases faster than does the temperature. Hence, the volume decreases.

7 • True. The kinetic energy of translation K for n moles of gas is directly proportional to the absolute temperature T of the gas ( )nkTK 2

3= .

8 • Determine the Concept We can use MRTv 3rms = to relate the temperature of a gas

to the rms speed of its molecules.

Express the dependence of the rms speed of the molecules of a gas on their absolute temperature:

MRTv 3

rms =

where R is the gas constant, M is the molar mass, and T is the absolute temperature.

molecules. theof speed rms thedouble toorder in quadrupled bemust re temperatu the, Because rms Tv ∝

9 • Picture the Problem The average kinetic energy of a molecule, as a function of the temperature, is given by kTK 2

3av = and the pressure, volume, and temperature of an ideal

gas are related according to .NkTPV =

Express the average kinetic energy of a molecule in terms of its temperature:

kTK 23

av =

Temperature and the Kinetic Theory of Gases

1315

From the ideal-gas law we have:

NkTPV =

Eliminate kT between these equations to obtain:

NPVK2

3av =

2. offactor aby increases ,constant at doubled is If avKVP

2. offactor aby increases ,constant at doubled is If avKPV

10 •• Picture the Problem We can express the rms speeds of the helium atoms and the methane molecules using .3rms MRTv =

Express the rms speed of the helium atoms: ( )

Herms

3HeMRTv =

Express the rms speed of the methane molecules: ( )

4CH4rms

3CHM

RTv =


( )( ) He

CH

4rms

rms 4

CHHe

MM

vv

=

Use Appendix C to find the molar masses of helium and methane:

( )( ) 2

g/mol4g/mol16

CHHe

4rms

rms ==vv

and correct. is )(b

11 • False. Whether the pressure changes also depends on whether and how the volume changes. In an isothermal process, the pressure can increase while the volume decreases and the temperature is constant.

12 • Determine the Concept For the Celsius scale, the ice point (0°C) and the boiling point of water at 1 atm (100°C) are more convenient than 273 K and 373 K; temperatures in roughly this range are normally encountered. On the Fahrenheit scale, the temperature of warm-blooded animals is roughly 100°F; this may be a more convenient reference than approximately 300 K. Throughout most of the world, the Celsius scale is the standard for nonscientific purposes.

Chapter 17

1316

*13 • Determine the Concept Because 107 >> 273, it does not matter.

14 • Determine the Concept The average speed of the molecules in an ideal gas depends on the square root of the kelvin temperature. Because ,av Tv ∝ doubling the temperature

while maintaining constant pressure increases the average speed by a factor of .2 correct. is )( d

15 • Determine the Concept From the ideal-gas law, we have .nRTPV = Halving both the temperature and volume of the gas leaves the pressure unchanged. correct. is )( b

16 • Determine the Concept The average translational kinetic energy of the molecules of an ideal gas is given by .2

323 nRTNkTK == The temperature of the ideal gas is related to

the pressure of the gas. correct. is )( d

17 • Determine the Concept The only conclusion we can draw from the information that the vessel contains equal amounts, by weight, of helium and argon is that the temperatures of the helium and argon molecules are the same. correct. is )( d

18 •• Determine the Concept The two rooms are in thermal equilibrium and, because they are connected, the air in each is at the same pressure. Because P = NkT/V, and the volume of each room is identical, NATA = NBTB, so the cooler room (A) has more air in it.

19 • Determine the Concept The rms speed of an ideal gas is given by MRTv 3rms = and

its average kinetic energy by .23

av kTK = Because the gases are at the same temperature,

their average kinetic energies are the same. Because ,3rms MRTv = the rms speeds

are inversely proportional to the square root of the molecular masses.

20 •• Determine the Concept The pressure is a measure of the change in momentum per second of a gas molecule on collision with the wall of the container. When the gas is heated, the average velocity, the average momentum, and pressure of the molecules increase.


1317

*21 •• Determine the Concept Because the temperature remains constant, the average speed of the molecules remains constant. When the volume decreases, the molecules travel less distance between collisions, so the pressure increases because the frequency of collisions increases.

22 •• Picture the Problem The average kinetic energies of the molecules are given by

( ) .23

av2

21

av kTmvK == Assuming that the room’s temperature distribution is uniform,

we can conclude that the oxygen and nitrogen molecules have equal average kinetic energies. Because the oxygen molecules are more massive, they must be moving slower than the nitrogen molecules. correct. is )( b

23 •• Determine the Concept The average molecular speed of He gas at 300 K is about 1.4 km/s, so a significant fraction of He molecules have speeds in excess of earth’s escape velocity (11.2 km/s). Thus, they "leak" away into space. Over time, the He content of the atmosphere decreases to almost nothing. Estimation and Approximation

*24 •• Picture the Problem Assuming the steam to be an ideal gas at a temperature of 373 K, we can use the ideal-gas law to estimate the pressure inside the test tube when the water is completely boiled away.

Using the ideal-gas law, relate the pressure inside the test tube to its volume and the temperature:

VNkTP =

Relate the number of particles N to the mass of water, its molar mass M, and Avogadro’s number NA:

ANM

Nm

=

Solve for N: M

NmN A=

Relate the mass of 1 mL of water to its density:

( )( ) g1m10kg/m10 3633 === −Vm ρ

Substitute for m, NA, and M (18 g/mol) and evaluate N:

( )

particles1035.3g/mol18

molparticles/10022.6g1

22

23

×=

×=N

Chapter 17

1318


( )( )( )

atm171

N/m101.01atm1N/m10172

m1010K373J/K10381.1particles1035.3

2525

36

2322

=

×××=

×××

= −

−

P

25 ••• Picture the Problem We can find the escape temperatures for the earth and the moon by equating, in turn, 0.15ve and vrms of O2 and H2. We can compare these temperatures to explain the absence from the earth’s upper atmosphere and from the surface of the moon.

(a) Express vrms for O2:

MRTv 3

rms =

where R is the gas constant, T is the absolute temperature, and M is the molar mass of oxygen.

Equate 0.15ve and vrms:

MRTgR 3215.0 earth =

Solve for T to obtain:

RMgRT

3045.0 earth= (1)

Evaluate T for O2: ( )( )

( )( )

K1061.3

kg/mol1032KJ/mol8.3143

m106.37m/s9.81045.0

3

3

62

×=

××

⋅×

=

−

T

(b) Substitute numerical values and evaluate T for H2:

( )( )( )

( )K225

kg/mol102KJ/mol8.3143

m106.37m/s9.81045.0

3

62

=

××

⋅×

=

−

T

(c) .atmosphereupper thefrom escape molecules H energetic

more theTherefore, escape. molecules H ,25or If

2

2atme51

rms TTvv ≥>


1319

(d) Express equation (1) at the surface of the moon:

( )

RMRg

RMRg

RMRgT

moonearth

moonearth61

moonmoon

0025.03

045.03

045.0

=

=

=

Substitute numerical values and evaluate T for O2:

( )( )( ) K164KJ/mol8.314

kg/mol1032m10738.1m/s9.810025.0 362

=⋅

××=

−

T

Substitute numerical values and evaluate T for H2:

( )( )( ) K3.10KJ/mol8.314

kg/mol102m10738.1m/s9.810025.0 362

=⋅

××=

−

T

present. themoon to theofformation thesince time theduring escaped have wouldH and O all then K, 1000ely approximatbeen have

wouldmoon on the re temperatu that theassume weIf

22

mospherewith an at

26 •• Picture the Problem We can use MRTv 3rms = to calculate the rms speeds of H2, O2,

and CO2 at 273 K and then compare these speeds to 20% of the escape velocity on Mars to decide the likelihood of finding these gases in the atmosphere of Mars. Express the rms speed of an atom as a function of the temperature:

MRTv 3

rms =

(a) Substitute numerical values and evaluate vrms for H2:

( )( )

km/s1.85

kg/mol102K273KJ/mol8.3143

3Hrms, 2

=

×⋅

= −v

(b) Evaluate vrms for O2: ( )( )

m/s614


3Orms, 2

=

×⋅

= −v

Chapter 17

1320

(c) Evaluate vrms for CO2: ( )( )

m/s933


3COrms, 2

=

×⋅

= −v

(d) Calculate 20% of vesc for Mars: ( ) km/s1km/s55

1esc5

1 === vv

.present be should ,Hnot but ,CO and OHfor than lessbut O and COfor an greater th is Because

222

2rms22rms vvv

*27 •• Picture the Problem We can use MRTv 3rms = to calculate the rms speeds of H2, O2,

and CO2 at 123 K and then compare these speeds to 20% of the escape velocity on Jupiter to decide the likelihood of finding these gases in the atmosphere of Jupiter. Express the rms speed of an atom as a function of the temperature:

MRTv 3

rms =

(a) Substitute numerical values and evaluate vrms for H2:

( )( )

km/s1.24


3Hrms, 2

=

×⋅

= −v

(b) Evaluate vrms for O2: ( )( )

m/s310


3Orms, 2

=

×⋅

= −v

(c) Evaluate vrms for CO2: ( )( )

m/s264


3COrms, 2

=

×⋅

= −v

(d) Calculate 20% of vesc for Jupiter: ( ) km/s12km/s0651

esc51 === vv

.Jupiteron found be should H and ,CO ,O ,H and ,CO ,Ofor an greater th is Because 222222rmsvv


1321

Temperature Scales 28 •

Picture the Problem We can convert both of these temperatures to the Fahrenheit scale and then express their difference to find the range of temperatures.

Solve the Fahrenheit-Celsius conversion equation for the Fahrenheit temperature:

°+= 32C59

F tt

Find the Fahrenheit equivalent of −12°C:

( ) °=°+°−= 4.10321259

Ft

Find the Fahrenheit equivalent of −7°C:

( ) F4.1932759

F °=°+°−=t

The difference between these two temperatures is the range on the Fahrenheit scale:

°=

°−°=

9.00F

F10.4F19.4Range

Remarks: We could take advantage of the fact that 9 F° = 5 C° to arrive at the aforementioned result in which the range of Celsius temperatures happens to be 5C°. If the temperature difference were other than 5C°, we could set up a proportion to quickly find the range on the Fahrenheit scale. 29 • Picture the Problem We can use the Fahrenheit-Celsius conversion equation to find this temperature on the Celsius scale. Convert 1945.4°F to the equivalent Celsius temperature:

( ) ( )C1063

324.194532 95

F95

C

°=

°−°=°−= tt

*30 • Picture the Problem We can use the Fahrenheit-Celsius conversion equation to express the temperature of the human body on the Celsius scale.

Convert 98.6°F to the equivalent Celsius temperature:

( ) ( )C0.37

326.9832 95

F95

C

°=

°−°=°−= tt

Chapter 17

1322

31 • Picture the Problem While we could use Equation 17-1 to relate the Celsius temperature to the length of the column of mercury in the thermometer, an alternative solution is to use the diagram to the right to set up a proportion that will relate the Celsius temperature to the calibration temperatures and to the lengths of the mercury column. Using the diagram, set up a proportion relating the temperatures to the lengths of the column of mercury:

0100

0tc

C0C001C0

LLLLt−−

=°−°

°−

Solve for and evaluate Lt: ( )

( )

( ) cm0.4100

cm0.20

cm0.4100

cm4.0cm0.24100

c

c

00100c

t

+°

=

+°−

=

+°−

=

t

t

LLLtL

(a) Substitute 22.0°C for tc and evaluate Lt:

( )( )

cm40.8

cm0.4100

cm0.20C0.22t

=

+°

°=L

(b) Substitute 25.4 cm for Lt and evaluate tc:

C107

100cm20

cm4.0cm25.4c

°=

°×−

=t

32 • Picture the Problem We can use the temperature conversion equations

°+= 32C59

F tt and K15.273C −= Tt to convert 107 K to the Fahrenheit and Celsius

temperatures.

Express the kelvin temperature in terms of the Celsius temperature:

K15.273C += tT


1323

(a) Solve for and evaluate tC:

K10

K273.15K10K15.2737

7C

≈

−=−= Tt

(b) Use the Celsius to Fahrenheit conversion equation to evaluate tF:

( ) F1080.132C10 7759

F °×≈°+°=t

33 • Picture the Problem While we could convert 77.35 K to a Celsius temperature and then convert the Celsius temperature to a Fahrenheit temperature, an alternative solution is to use the diagram to the right to set up a proportion for the direct conversion of the kelvin temperature to its Fahrenheit equivalent.

Use the diagram to set up the proportion: K273.15K373.15

K77.35K273.15F32F212

F32 F

−−

=°−°

−° t

or

100195

F801F32 F =

°−° t

Solve for and evaluate tF: F319F180

100195F32F °−=°×−°=t

34 • Picture the Problem We can use the fact that, for a constant-volume gas thermometer, the pressure and absolute temperature are directly proportional to calibrate the given thermometer; i.e., to find the constant of proportionality relating P and T. We can then use this equation to find the temperature corresponding to a given pressure or the pressure corresponding to a given temperature. Express the direct proportionality between the pressure and the temperature:

CTP = where C is a constant.

Use numerical values to evaluate C:

atm/K101.464K273.15

atm0.400

3−×=

==TPC

Chapter 17

1324

Substitute to obtain: ( )TP atm/K10464.1 3−×= (1)

or ( )PT K/atm9.682= (2)

(a) Use equation (2) to find the temperature:

( )( )K68.3

atm0.1K/atm682.9

=

=T

(b) Use equation (1) to find the boiling point of sulfur:

( )( )

atm05.1

K15.2736.444atm/K101.464 3

=

+××= −P

*35 • Picture the Problem We can use the information that the thermometer reads 50 torr at the triple point of water to calibrate it. We can then use the direct proportionality between the absolute temperature and the pressure to either the pressure at a given temperature or the temperature for a given pressure. Using the ideal-gas temperature scale, relate the temperature to the pressure:

( )P

PPP

T

K/torr463.5torr50

K16.273K16.2733

=

==

(a) Solve for and evaluate P when T = 300 K:

( )( )( )

torr9.54

K300torr/K1830.0torr/K1830.0

=

== TP

(b) Find T when the pressure is 678 torr: ( )( )

K3704

torr678K/torr463.5

=

=T

36 • Picture the Problem We can use the equation for the ideal-gas temperature scale to express the temperature measured by this thermometer in terms of its pressure and the given data to calibrate the thermometer. Write the equation for the ideal-gas temperature scale:

PP

T3

K16.273= 17-4


1325

(a) Solve for and evaluate the thermometer’s triple-point pressure:

( )

torr0.22

torr30K373

K273.16K16.2733

=

== PT

P

(b) Substitute for P3 in Equation 17-4: ( )

K17.2

torr175.0torr0.22

K16.273torr0.22

K16.273

=

== PT

37 • Picture the Problem We can find the temperature at which the Fahrenheit and Celsius scales give the same reading by setting tF = tC in the temperature-conversion equation. Set tF = tC in ( )°−= 32F9

5C tt : ( )°−= 32F9

5F tt

Solve for and evaluate tF: F0.40C0.40FC °−=°−== tt

Remarks: If you’ve not already thought of doing so, you might use your graphing calculator to plot tC versus tF and tF = tC (a straight line at 45°) on the same graph. Their intersection is at (−40, −40). 38 • Picture the Problem We can use the Celsius-to-absolute conversion equation to find 371 K on the Celsius scale and the Celsius-to-Fahrenheit conversion equation to find the Fahrenheit temperature corresponding to 371 K. Express the absolute temperature as a function of the Celsius temperature:

K15.273C += tT

Solve for and evaluate tC:

C97.9K273.15K371

K15.273C

°=−=

−= Tt

Use the Celsius-to-Fahrenheit conversion equation to find tF:

( )F208

329.9732 59

C59

F

°=

°+°=°+= tt

39 • Picture the Problem We can use the Celsius-to-absolute conversion equation to find 90.2 K on the Celsius scale and the Celsius-to-Fahrenheit conversion equation to find the

Chapter 17

1326

Fahrenheit temperature corresponding to 90.2 K.

Express the absolute temperature as a function of the Celsius temperature:

K15.273C += tT

Solve for and evaluate tC:

C183K273.15K2.09

K15.273C

°−=−=

−= Tt

Use the Celsius-to-Fahrenheit conversion equation to find tF:

( )F297

3218332 59

C59

F

°−=

°+°−=°+= tt

40 •• Picture the Problem We can use the diagram to the right to set up proportions that will allow us to convert temperatures on the Réaumur scale to Celsius and Fahrenheit temperatures.

Referring to the diagram, set up a proportion to convert temperatures on the Réaumur scale to Celsius temperatures:

R0R80R0

C0C100C0 RC

°−°°−

=°−°

°− tt


RC tt= or RC 25.1 tt =

Referring to the diagram, set up a proportion to convert temperatures on the Réaumur scale to Fahrenheit temperatures:

R0R80R0

F32F212F32 RF

°−°°−

=°−°

°− tt


32 RF tt=

− or 32R4

9F += tt

*41 ••• Picture the Problem We can use the temperature dependence of the resistance of the thermistor and the given data to determine R0 and B. Once we know these quantities, we


1327

can use the temperature-dependence equation to find the resistance at any temperature in the calibration range. Differentiation of R with respect to T will allow us to express the rate of change of resistance with temperature at both the ice point and the steam point temperatures.

(a) Express the resistance at the ice point as a function of temperature of the ice point:

K27307360 BeR=Ω (1)

Express the resistance at the steam point as a function of temperature of the steam point:

K3730153 BeR=Ω (2)


K373K27310.481537360 BBe −==

ΩΩ

Solve for B by taking the logarithm of both sides of the equation:

1K3731

27311.48ln −⎟

⎠⎞

⎜⎝⎛ −= B

and

K1094.3K

3731

2731

1.48ln 3

1×=

⎟⎠⎞

⎜⎝⎛ −

=−

B

Solve equation (1) for R0 and substitute for B:

( )

( )Ω×=

Ω=

Ω=Ω

=

−

×−

−

3

K273K1094.3

K273K2730

1097.3

7360

73607360

3

e

ee

R BB

(b) From (a) we have: ( ) TeR K1094.33 3

1097.3 ×− Ω×=

Convert 98.6°F to kelvins to obtain: K310=T


Ω=

Ω×= ×−

k31.1

1097.3 K310K1094.33 3

eR

(c) Differentiate R with respect to T to obtain:

( )

202

00

TRBeR

TB

TB

dTdeReR

dTd

dTdR

TB

TBTB

−=−

=

⎟⎠⎞

⎜⎝⎛==

Chapter 17

1328

Evaluate dR/dT at the ice point:

( )( )( )

K/389

K16.273K1094.37360

2

3

pointice

Ω−=

×Ω−=⎟

⎠⎞

⎜⎝⎛

dTdR

Evaluate dR/dT at the steam point:

( )( )( )

K/33.4

K16.373K1094.3153

2

3

pointsteam

Ω−=

×Ω−=⎟

⎠⎞

⎜⎝⎛

dTdR

(d) es.temperatur

lowerat y sensitivitgreater hasit i.e.,sensitive; more isr thermistoThe

The Ideal-Gas Law 42 •

Picture the Problem Let the subscript 1 refer to the gas at 50°C and the subscript 2 to the gas at 100°C. We can apply the ideal-gas law for a fixed amount of gas to find the ratio of the final and initial volumes.

Apply the ideal-gas law for a fixed amount of gas:

1

11

2

22

TVP

TVP

=

or, because P2 = P1,

1

2

1

2

TT

VV

=

Substitute numerical values and evaluate V2/V1:

( )( ) 15.1

K5015.273K10015.273

1

2 =++

=VV

43 •

Picture the Problem We can use the ideal-gas law to find the number of moles of gas in the vessel and the definition of Avogadro’s number to find the number of molecules.

Apply the ideal-gas law to the gas: nRTPV =

Solve for the number of moles of gas in the vessel: RT

PVn =


1329

Substitute numerical values and evaluate n:

( )( )( )( )

mol1.79

K273Katm/molL108.206L10atm4

2

=

⋅⋅×= −n

Relate the number of molecules N in the gas in terms of the number of moles n:

AnNN =

Substitute numerical values and evaluate N:

( )( ) molecules101.08molmolecules/106.022mol1.79 2423 ×=×=N

44 •• Picture the Problem We can use the ideal-gas law to relate the number of molecules in the gas to its pressure, volume, and temperature. Solve the ideal-gas law for the number of molecules in a gas as a function of its pressure, volume, and temperature:

kTPVN =

Substitute numerical values and evaluate N:

( )( )( )( )( )

8

23

368

1022.3

K300J/K101.381m10Pa/torr133.32torr10

×=

×= −

−−

N

45 •• Picture the Problem The pictorial representation to the right, in which T0 represents absolute zero, summarizes the information concerning the temperatures and pressures we are given. We know, from the ideal-gas law, that the pressure of a fixed volume of gas is proportional to its absolute temperature. We can use the diagram to set up a proportion relating the temperatures and pressures that we can solve for T0. Apply the ideal-gas law to obtain: klads

Tglipsklads

Tglips7.8

105.12

22 00 −−=

−

Solve for T0 to obtain: glipsT 2.830 −=

Chapter 17

1330

Remarks: Because the gas is ideal, its pressure is directly proportional to its temperature. Hence, a graph of P versus T will be linear and the linear equation relating P and T can be solved for the temperature corresponding to zero pressure. 46 •• Picture the Problem Let the subscript 1 refer to the tires when their pressure is 180 kPa and the subscript 2 to conditions when their pressure is 245 kPa. Assume that the air in the tires behaves as an ideal gas. Then, we can apply the ideal-gas law for a fixed amount of gas to relate the temperatures to the pressures and volumes of the tires. (a) Apply the ideal-gas law for a fixed amount of gas to the air in the tires:

1

11

2

22

TVP

TVP

= (1)

Solve for T2: 1

212 P

PTT = because V1 = V2.


( )

C87.7

K360.7kPa180kPa245K2652

°=

==T

(b) Use equation (1) with V2 = 1.07 V1. Solve for T2: ⎟⎟

⎠

⎞⎜⎜⎝

⎛== 1

1

21

11

222 07.1 T

PPT

VPVPT

Substitute numerical values and evaluate T2:

( ) C113K385.9K7.36007.12 °===T

47 •• Picture the Problem We can apply the ideal-gas law to find the number of moles of air in the room as a function of the temperature.

(a) Use the ideal-gas law to relate the number of moles of air in the room to the pressure, volume, and temperature of the air:

RTPVn = (1)


( )( )( )( )

mol103.66

K300KJ/mol8.314m90kPa101.3

3

3

×=

⋅=n


1331

(b) Letting n′ represent the number of moles in the room when the temperature rises by 5 K, express the number of moles of air that leave the room:

n'nn −=∆

Apply the ideal-gas law to obtain: 'RT

PVn' = (2)

Divide equation (2) by equation (1) to obtain: T'

Tnn'

= and T'Tnn' =

Substitute for n′ to obtain:

⎟⎠⎞

⎜⎝⎛ −=−=∆

T'Tn

T'Tnnn 1

Substitute numerical values and evaluate ∆n: ( )

mol60.0

K305K3001mol103.66 3

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−×=∆n

*48 •• Picture the Problem Let the subscript 1 refer to helium gas at 4.2 K and the subscript 2 to the gas at 293 K. We can apply the ideal-gas law to find the volume of the gas at 4.2 K and a fixed amount of gas to find its volume at 293 K.

(a) Apply the ideal-gas law to the helium gas and solve for its volume:

1

11 P

nRTV =


( )( )

( )n

nV

mol/L3447.0atm1

K4.2Katm/molL0.082061

=

⋅⋅=

Find the number of moles in 10 g of helium:

mol2.5g/mol4

g10==n

Substitute for n to obtain:

( )( )L862.0

mol5.2mol/L3447.01

=

=V

(b) Apply the ideal-gas law for a fixed amount of gas and solve for 1

11

2

22

TVP

TVP

=

Chapter 17

1332

the volume of the helium gas at 293 K:

and, because P1 = P2,

11

22 V

TTV =

Substitute numerical values and evaluate V2:

( ) L60.1L0.862K4.2K293

2 ==V

49 •• Picture the Problem Because the helium is initially in the liquid state, its temperature must be 4.2 K. Let the subscript 1 refer to helium gas at 4.2 K and the subscript 2 to the gas at 293 K. We can apply the ideal-gas law for a fixed volume of gas to relate the pressure at 293 K to the pressure at 4.2 K and use the ideal-gas law to find the pressure at 4.2 K.


1

11

2

22

TVP

TVP

=

Solve for its pressure at 293 K: 1

21

12

2112 T

TPTVTVPP == (1)

because V1 = V2

Apply the ideal-gas law to the helium gas at 4.2 K and solve for its pressure:

1

11 V

nRTP =


( )( )

( )n

nP

mol/atm05744.0L6

K4.2Katm/molL0.082061

=

⋅⋅=

Find the number of moles in 10 g of helium:

mol2.5g/mol4

g10==n

Substitute for n to obtain:

( )( )atm1436.0

mol5.2mol/atm05744.01

==P

Substitute in equation (1) and evaluate P2:

( ) atm10.0K4.2K293atm0.14362 ==P

*50 •• Picture the Problem Let the subscript 1 refer to the tire when its temperature is 20°C and the subscript 2 to conditions when its temperature is 50°C. We can apply the ideal-


1333

gas law for a fixed amount of gas to relate the temperatures to the pressures of the air in the tire.

(a) Apply the ideal-gas law for a fixed amount of gas and solve for pressure at the higher temperature:

1

11

2

22

TVP

TVP

= (1)

and

11

22 P

TTP =

because V1 = V2


( )

kPa323

kPa101kPa200K293K323

2

=

+=P

and

kPa231

kPa101kPa332gauge2,

=

−=P

(b) Solve equation (1) for P2 with V2 = 1.1 V1 and evaluate P2:

( ) ( )

kPa302

kPa101kPa200K2931.1

K323

112

212

=

+=

= PTVTVP

and kPa201kPa101kPa302gauge2, =−=P

51 •• Picture the Problem Let

2Nρ and20ρ be the number densities (i.e., the number of

particles per unit volume) of N2 and O2, respectively. We can express the density of air in terms of the densities of nitrogen and oxygen and their number densities as .

2222 00NNair ρρρ mm += By applying the ideal-gas law, we can find the number density of air and, using the given composition of air, calculate the number densities of nitrogen and oxygen. Finally, we can find the masses of nitrogen and oxygen molecules from their atomic masses. Knowing ,

2Nρ ,2Oρ ,

2Nm and ,2Om we can calculate ρair.

Express the density of air in terms of the densities of nitrogen and oxygen:

2222 00NNair ρρρ mm += (1)

Using the ideal-gas law, relate the number density of air N/V to its temperature and pressure:

NkTPV = and kTP

VN

=

Chapter 17

1334

Substitute numerical values and evaluate the number density of air: ( )( )

325

23

25

m1046.2K297J/K10381.1

N/m1001.1

−

−

×=

××

=VN

Because air is approximately 74% N2 and 26% O2:

( )325

325N

m1082.1

m1046.274.074.02

−

−

×=

×==VNρ

and

( )324

325O

m1040.6

m1046.226.026.02

−

−

×=

×==VNρ

Calculate the masses of N2 and O2 molecules:

( )( )kg1065.4

kg/u10660.1u2826

27N 2

−

−

×=

×=m

and ( )( )

kg1031.5

kg/u10660.1u3226

27O2

−

−

×=

×=m

Substitute in equation (1) and evaluate ρair:

( )( )( )( )

3

32426

32526air

kg/m19.1

m1040.6kg1031.5

m1082.1kg1065.4

=

××+

××=−−

−−ρ

52 •• Picture the Problem Let the subscript 1 refer to the conditions at the bottom of the lake and the subscript 2 to the surface of the lake and apply the ideal-gas law for a fixed amount of gas.


1

11

2

22

TVP

TVP

=

Solve for the volume of the bubble just before it breaks the surface: 21

1212 PT

PTVV =

Find the pressure at the bottom of the lake:

( )( )( )kPa7.493

m40m/s9.81kg/m10kPa01.31

233

atm1

=+

=+= ghPP ρ


1335

Substitute numerical values and evaluate V2:

( )( )( )( )( )3

32

cm4.78

kPa101.3K278kPa493.7K298cm15

=

=V

53 •• Picture the Problem Assume that the volume of the balloon is not changing. Then the air inside and outside the balloon must be at the same pressure of about 1 atm. The contents of the balloon are the air molecules inside it. We can use Archimedes principle to express the buoyant force on the balloon and we can find the weight of the air molecules inside the balloon Express the net force on the balloon and its contents:

balloon theinsideair net wBF −= (1)

Using Archimedes principle, express the buoyant force on the balloon:

gmwB fluid displacedfluid displaced == or

gVB balloonoρ= where ρo is the density of the air outside the balloon.

Express the weight of the air inside the balloon:

gVw ballooniballoon theinsideair ρ= where ρi is the density of the air inside the balloon.

Substitute in equation (1) for B and wair inside the balloon to obtain: ( ) gV

gVgVF

balloonio

ballooniballoononet

ρρρρ

−=−=

(2)

Express the densities of the air molecules in terms of their number densities, molecular mass, and Avogadro’s number:

⎟⎠⎞

⎜⎝⎛=

VN

NM

A

ρ

Using the ideal-gas law, relate the number density of air N/V to its temperature and pressure:

NkTPV = and kTP

VN

=

Substitute to obtain: ⎟

⎠⎞

⎜⎝⎛=

kTP

NM

A

ρ


gVTTkN

MPF balloonioA

net11⎟⎟⎠

⎞⎜⎜⎝

⎛−=

Chapter 17

1336

Assuming that the average molecular weight of air is 29 g/mol, substitute numerical values and evaluate Fnet:

( )( )( )( )

( )( )N56.2

m/s81.9m5.1

K3481

K2971

J/K10381.1molparticles/10022.6N/m1001.1g/mol29

23

2323

25

net

=

×

⎟⎟⎠

⎞⎜⎜⎝

⎛−

×××

= −F

54 ••• Picture the Problem We can find the number of moles of helium gas in the balloon by applying the ideal-gas law to relate n to the pressure, volume, and temperature of the helium and Archimedes principle to find the volume of the helium. In part (b), we can apply the result of Problem 13-95 to relate atmospheric pressure to altitude and use the ideal-gas law to determine the pressure of the gas when the balloon is fully inflated. In part (c), we’ll find the net force acting on the balloon at the altitude at which it is fully inflated in order to decide whether it can rise to that altitude.

(a) Apply the ideal-gas law to the helium in the balloon and solve for n:

RTPVn = (1)

Relate the net force on the balloon to its weight:

N30HeloadskinB =−−− wwwF

Use Archimedes principle to express the buoyant force on the balloon in terms of the volume of the balloon:

Vg

wF

air

airdisplacedB

ρ=

=


N30Heloadskinair =−−− VgwwVg ρρ

Solve for the volume of the helium: ( )g

wwVHeair

loadskinN30ρρ −++

=

Substitute numerical values and evaluate V: ( )

( )3

2

33

m39.17m/s9.811

kg/m0.179kg/m1.293N110N50N30

=

×

−++

=V


1337

Substitute numerical values in equation (1) and evaluate n: ( )( )

( )( )mol776

K273Katm/molL108.206m10

L1m17.39atm1

2

333

=

⋅⋅×

⎟⎟⎠

⎞⎜⎜⎝

⎛

= −

−

n

(b) Using the result of Problem 13-95, express the variation in atmospheric pressure with altitude:

( ) 00

hhePhP −=

where h0 = 7.93 km

Solve for h: ( )⎥⎦

⎤⎢⎣

⎡=

hPPhh 0

0 ln (2)

Neglecting changes in temperature with elevation, apply the ideal-gas law to find the pressure at which the balloon’s volume will be 32 m3:

( )( )( ) atm543.0

m10L1m32

K273Katm/molL108.206mol776

333

2

=×

⋅⋅×==

−

−

VnRTP

Substitute in equation (2) and evaluate h: ( ) km84.4

atm0.543atm1lnkm93.7 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=h

(c) Express the condition that must be satisfied if the balloon is to reach its fully inflated altitude:

0totBnet ≥−= wFF (3)

Express wtot:

He

He

Heskinloadtot

N16050110w

wNNwwww

+=++=

++=

Express the weight of the helium: Vgw HeHe ρ=

Substitute for wHe and evaluate wtot:

( )( )( )

N5.190m/s9.81

m17.38kg/m0.179N601

N160

2

33Hetot

=×

+=

+= Vgw ρ

Chapter 17

1338

Determine the buoyant force on the balloon at h = 4.84 km:

VgF hair,B ρ= (4)

Express the dependence of the density of the air on atmospheric pressure:

air

air,

0 ρρ h

PP= (5)

or

air0

air, ρρPP

h =

Substitute and evaluate FB:

( )( )( )N9.219

m/s9.81m23kg/m1.293543.0

2

33

air0

B

=×

=

= VgPPF ρ

Substitute in equation (3) and evaluate Fnet:

0N29.4N190.5N19.92net ≥=−=F

inflated.fully isit at which altitude than thehigher rise llballoon wi the,0 Because net >F

(d) The balloon will rise until the net force acting on it is zero. Because the buoyant force depends on the density of the air, the balloon will rise until the density of the air has decreased sufficiently for the buoyant force to just equal the total weight of the balloon. Substitute equation (5) in equation (2) to obtain:

h

hhair,

air0 ln

ρρ

=

Using equation (4), find the density of the air such that FB = 190.5 N: ( )( )

3

23B

air,

kg/m0.6068m/s9.81m32

N190.5

=

==VgF

hρ

Substitute numerical values and evaluate h: ( )

km00.6

kg/m6068.0kg/m293.1lnkm93.7 3

3

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=h


1339

Kinetic Theory of Gases *55 • Picture the Problem We can express the rms speeds of argon and helium atoms by combining nRTPV = and MRTv 3rms = to obtain an expression for vrms in terms of

P, V, and M.

Express the rms speed of an atom as a function of the temperature:

MRTv 3

rms =

From the ideal-gas law we have: n

PVRT =


nMPVv 3

rms =

(a) Substitute numerical values and evaluate vrms for an argon atom:

( ) ( )( )( )( )( ) m/s276

kg/mol1040mol1m10kPa/atm101.3atm103Ar 3

33

rms =×

= −

−

v

(b) Substitute numerical values and evaluate vrms for a helium atom:

( ) ( )( )( )( )( ) m/s872

kg/mol104mol1m10kPa/atm101.3atm103He 3

33

rms =×

= −

−

v

56 • Picture the Problem We can express the total translational kinetic energy of the oxygen gas by combining nRTK 2

3= and the ideal-gas law to obtain an expression for K in terms

of the pressure and volume of the gas.

Relate the total translational kinetic energy of translation to the temperature of the gas:

nRTK 23=

Using the ideal-gas law, substitute for nRT to obtain:

PVK 23=


( )( ) J152m10kPa101.3 3323 == −K

Chapter 17

1340

57 • Picture the Problem Because we’re given the temperature of the hydrogen atom and know its molar mass, we can find its rms speed using MRTv 3rms = and its average

kinetic energy from .23

av kTK =

Relate the rms speed of a hydrogen atom to its temperature and molar mass:

MRTv 3

rms =

Substitute numerical values and evaluate vrms:

( )( )

km/s499


3

7

rms

=

⋅= −v

Express the average kinetic energy of the hydrogen atom as a function of its temperature:

kTK 23

av =

Substitute numerical values and evaluate Kav:

( )( )J102.07

K10J/K101.38116

72323

av

−

−

×=

×=K

*58 • Picture the Problem Because there are 6 squared terms in the expression for the total energy of an atom in this model, we can conclude that there are 6 degrees of freedom. Because the system is in equilibrium, we can conclude that there is energy of kT2

1 per molecule or RT2

1 per mole associated with each degree of freedom.

Express the average energy per atom in the solid in terms of its temperature and the number of degrees of freedom:

( ) ( ) kTkTkTNE 36atom 2

121av ===

Relate the total energy of one mole to its temperature and the number of degrees of freedom:

( ) ( ) RTRTRTNE 36mole 2

121tot ===

59 •

Picture the Problem We can combine22

1dnvπ

λ = and nRTPV = to express the

mean free path for a molecule in an ideal gas in terms of the pressure and temperature.


1341

Express the mean free path of a molecule in an ideal gas:

221

dnvπλ =

where VnNVNnv A==

Solve the ideal-gas law for the volume of the gas:

PnRTV =

Substitute in our expression for nv to obtain: kT

PPnRTnNnv == A

Substitute in the expression for the mean free path to obtain: 22 dP

kTπ

λ =

60 •• Picture the Problem We can find the collision time from the mean free path and the average (rms) speed of the helium molecules. We can use the result of Problem 43 to find the mean free path of the molecules and MRTv 3rms = to find the average speed of the

molecules.

Express the collision time in terms of the mean free path for and the average speed of a helium molecule:

avvλτ = (1)

Use the result of Problem 43 to express the mean free path of the gas:

22 dPkTπ

λ =

Substitute numerical values and evaluate λ:

( )( )( ) ( )

m101.332m10Pa1072K300J/K101.381

9

21011

23

×=

×

×=

−−

−

πλ

Express the average speed of the molecules: M

RTv 3rms =


( )( )

m/s101.368


3

3rms

×=

×⋅

= −v

Chapter 17

1342

Substitute in equation (1) and evaluate τ:

s109.74m/s101.368m101.332 5

3

9

×=××

=τ

*61 •• Picture the Problem We can use kTK 2

3= and ANMghmghU ==∆ to express the

ratio of the average kinetic energy of a molecule of the gas to the change in its gravitational potential energy if it falls from the top of the container to the bottom.

Express the average kinetic energy of a molecule of the gas as a function of its temperature:

kTK 23=

Letting h represent the height of the container, express the change in the potential energy of a molecule as it falls from the top of the container to the bottom:

ANMghmghU ==∆

Express the ratio of K to ∆U and simplify to obtain:

MghkTN

NMgh

kTUK

23 A

A

23

==∆

Substitute numerical values and evaluate K/∆U:

( )( )( )( )( )( )

423

2323

1095.7m15.0m/s81.9kg10322

K300J/K10381.110022.63×=

×××

=∆ −

−

UK

The Distribution of Molecular Speeds 62 •• Picture the Problem Equation 17-37 gives the Maxwell-Boltzmann speed distribution. Setting its derivative with respect to v equal to zero will tell us where the function’s extreme values lie. Differentiate Equation 17-37 with respect to v:

kTmv

kTmv

ekTmvv

kTm

evkTm

dvd

dvdf

2323

2223

2

2

22

4

24

−

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

π

π


1343

Set df/dv = 0 for extrema and solve for v: 02

3

=−kTmvv ⇒

mkTv 2

=

Examination of the graph of f(v) makes it

clear that this extreme value is, in fact, a maximum. See Figure 17-16 and note that it is concave downward at .2 mkTv =

Remarks: An alternative to the examination of f(v) in order to conclude that

mkTv 2= maximizes the Maxwell-Boltzmann speed distribution function is to

show that d2f/dv2 < 0 at .mkTv 2=

*63 •• Picture the Problem We can show that f(v) is normalized by using the given integral to integrate it over all possible speeds.

Express the integral of Equation 17-37:

( ) ∫∫∞

−∞

⎟⎠⎞

⎜⎝⎛=

0

2223

0

2

24 dvev

kTmdvvf kTmv

π

Let kTma 2= to obtain:

( ) ∫∫∞

−∞

=0

223

0

24 dvevadvvf av

π

Use the given integral to obtain:

( ) 14

4 2323

0

=⎟⎟⎠

⎞⎜⎜⎝

⎛= −

∞

∫ aadvvf ππ

i.e., f(v) is normalized.

64 •• Picture the Problem In Problem 63 we showed that f(v) is normalized. Hence we can

evaluate vav using ( )∫∞

0

dvvvf .

Express the average speed of the molecules in the gas:

( )

∫

∫∞

−

∞

⎟⎠⎞

⎜⎝⎛=

=

0

2323

0av

2

24 dvev

kTm

dvvvfv

kTmv

π

Chapter 17

1344

Substitute :2kTma = ∫∞

−=0

323av

24 dvevav av

π

Use the given integral to obtain:

mkT

aaav

22

122

4 223

av

π

ππ

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

*65 •• Picture the Problem Choose a coordinate system in which downward is the positive direction. We can use a constant-acceleration equation to relate the fall distance to the initial velocity of the molecule, the acceleration due to gravity, the fall time, and

mkTv 3rms = to find the initial velocity of the molecules.

(a) Using a constant-acceleration equation, relate the fall distance to the initial velocity of a molecule, the acceleration due to gravity, and the fall time:

221

0 gttvy += (1)

Express the rms speed of the atom to its temperature and mass:

mkTv 3

rms =


( )( )( )( )

m/s1092.5

kg/u10660.1u47.85nK120J/K10381.13

3

27

23

rms

−

−

−

×=

××

=v

Letting vrms = v0, substitute in equation (1) to obtain:

( ) ( ) 22213 m/s81.9m/s1092.5m1.0 tt +×= −

Solve this equation to obtain: s142.0=t

(b) If the atom is initially moving upward:

m/s1092.5 30rms

−×−== vv


( )( ) 22

21

3

m/s81.9m/s1092.5m1.0

tt

+

×−= −


1345

Solve this equation to obtain: s143.0=t

General Problems 66 • Picture the Problem We can use MRTv 3rms = to relate the temperature of the H2

molecule to its rms speed.

Relate the rms speed of the molecule to its temperature:

MRTv 3

rms =

Solve for the temperature: R

MvT3

2rms=


( )( )( )K8.79

KJ/mol8.3143m/s331kg/mol102 23

=

⋅×

=−

T

67 •• Picture the Problem We can use the ideal-gas law to find the initial temperature of the gas and the ideal-gas law for a fixed amount of gas to relate the volumes, pressures, and temperatures resulting from the given processes. (a) Apply the ideal-gas law to express the temperature of the gas: nR

PVT =


( )( )( )( ) K122

KJ/mol8.314mol1m1010kPa101.3 33

=⋅

×=

−

T

(b) Use the ideal-gas law for a fixed amount of gas to relate the temperatures and volumes:

2

22

1

11

TVP

TVP

=


2

2

1

1

TV

TV

=

Solve for and evaluate T2: ( ) K244K122211

22 === T

VVT

(c) Use the ideal-gas law for a fixed amount of gas to relate the 2

22

1

11

TVP

TVP

=

Chapter 17

1346

temperatures and pressures: or, because V1 = V2,

2

2

1

1

TP

TP=

Solve for T2:

11

22 P

TTP =


( ) atm1.43atm1K244K350

2 ==P

68 •• Picture the Problem We can use the definition of pressure to express the net force on each wall of the box in terms of its area and the pressure differential between the inside and the outside of the box. We can apply the ideal-gas law for a fixed amount of gas to find the pressure inside the box.

Using the definition of pressure, express the net force on each wall of the box:

( )outsideinside PPAPAF

−=∆=

Use the ideal-gas law for a fixed amount of gas to relate the initial and final pressures of the gas:

2

22

1

11

TVP

TVP

=

or, because V1 = V2,

2

2

1

1

TP

TP=

Solve for and evaluate Pinside: ( )

kPa135.1

kPa101.3K300K400

11

2inside2

=

=== PTTPP

Substitute and evaluate F: ( ) ( )

kN1.35

kPa101.3kPa135.1m0.2 2

=

−=F

*69 •• Picture the Problem We can use the molar mass of water to find the number of moles in 2 L of water. Because there are two hydrogen atoms in each molecule of water, there must be as many hydrogen molecules in the gas formed by electrolysis as there were molecules of water and, because there is one oxygen atom in each molecule of water, there must be half as many oxygen molecules in the gas formed by electrolysis as there were molecules of water.


1347

Express the electrolysis of water into H2 and O2:

( ) ( ) ( )221

22 OHOH nnn +→

Express the number of moles in 2 L of water:

( ) mol111g/mol18

g2000OH2 ==n

Because there is one hydrogen atom for each water molecule:

( ) mol111H2 =n

Because there are two oxygen atoms for each water molecule:

( ) ( ) ( )mol5.55

mol111OHO 21

221

2

=

== nn

70 •• Picture the Problem The diagram shows the cylinder before removal of the membrane. We’ll assume that the gases are at the same temperature. The approximate location of the center of mass (CM) is indicated. We can find the distance the cylinder moves by finding the location of the CM after the membrane is removed.

Express the distance the cylinder will move in terms of the movement of the center of mass when the membrane is removed:

beforecm,aftercm, xxx −=∆

Apply the ideal-gas law to both collections of molecules to obtain:

( )kTnVP 2NN N22=

and ( )kTnVP 2OO O

22=


( )( )2

2

O

N

ON

2

2

nn

PP

=

or, because 22 ON 2PP = ,

( )( )2

2

O

O

ON2

2

2

nn

PP

= ⇒ n(N2) = 2n(O2)

Express the mass of O2 in terms of its molar mass and the number of moles of oxygen:

m(O2) = n(O2)M(O2)

Chapter 17

1348

Express the mass of N2 in terms of its molar mass and the number of moles of nitrogen:

m(N2) = 2n(O2)M(N2).

Using its definition, express the center of mass before the membrane is removed:

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( )( ) ( )22

Ocm,2Ncm,2

2222

Ocm,22Ncm,22

2222

Ocm,22Ncm,22beforecm,

ON2ON2

OONO2OONO2

OONNOONN

22

22

22

MMxMxM

MnMnxMnxMn

MnMnxMnxMn

m

mxx

ii

iii

++

=

++

=

++

==∑∑

Substitute numerical values and evaluate xcm,before:

( )( ) ( )( )( ) cm17.27

g32g282g32cm30g28cm102

beforecm, =++

=x

Locate the center of mass after the membrane is removed:

( )( ) ( )( )( )

cm0.02g32g282

g32cm20g28cm202aftercm,

=++

=x


cm2.73

cm17.27cm20.00

=

−=∆x

left. the tocm 2.73 movedcylinder theright, the tomoved mass ofcenter theand process thisduring conserved bemust momentum Because

71 •• Picture the Problem We can apply the ideal-gas law to the two processes to find the number of moles of hydrogen in terms of the number of moles of nitrogen in the gas. Using the definition of molar mass, we can relate the mass of each gas to the number of moles of each gas and their molar masses.

Apply the ideal-gas law to the first case:

( ) ( )[ ] 1221 HN2 RTnnVP +=

Apply the ideal-gas law to the ( ) ( )[ ] 1221 2H2N23 RTnnVP +=


1349

second case:

Divide the second of these equations by the first and simplify to express n(H2) in terms of n(N2):

( ) ( )22 N2H nn = (1)

Relate the mN to n(N2): ( ) ( )( )( )g/mol28N

NN

2

22N

nMnm

==

and

( )g/mol28

N N2

mn =

Relate the mH to n(H2): ( ) ( )

( )( )g/mol2HHH

2

22H

nMnm

==

and

( )g/mol2

H H2

mn =

Substitute in equation (1) and solve for mN: g/mol28

2g/mol2

NH mm= ⇒ HN 7mm =

*72 •• Picture the Problem Initially, we have 3P0V = n0RT0. Later, the pressures in the three vessels, each of volume V, are still equal, but the number of moles is not. The total number of moles, however, is constant and equal to the number of moles in the three vessels initially. Applying the ideal-gas law to each of the vessels will allow us to relate the number of moles in each to the final pressure and temperature. Equating this sum n0 will leave us with an equation in P′ and P0 that we can solve for P′.

Relate the number of moles of gas in the system in the three vessels initially to the number in each vessel when the pressure is P′:

3210 nnnn ++=

Relate the final pressure in the first vessel to its temperature and solve for n1:

( )V

TRnP' 01 2= ⇒

01 2RT

P'Vn =

Relate the final pressure in the second vessel to its temperature and solve for n2:

( )V

TRnP' 02 3= ⇒

02 3RT

P'Vn =

Chapter 17

1350

Relate the final pressure in the third vessel to its temperature and solve for n3:

VRTnP' 03= ⇒

03 RT

P'Vn =


⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛ ++=

++=

00

0000

6111

31

21

32

RTP'V

RTP'V

RTP'V

RTP'V

RTP'Vn

Express the number of moles in the three vessels initially in terms of the initial pressure and total volume:

( )0

00

3RT

VPn =

Equate the two expressions for n0 and solve for P′ to obtain: 011

18 PP' =

73 •• Picture the Problem We can use the ideal-gas temperature scale to relate the temperature of the boiling substance to its pressure and the pressure at the triple point. If we assume a linear relationship between P/P3 and P3, we can calibrate this equation using the data from any two (or all) of the temperature measurements and then extrapolate this equation to zero gas pressure to find the ideal-gas temperature of the boiling substance.

Using the ideal-gas temperature scale, relate the temperature of the boiling substance to its pressure and the pressure at the triple point:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

3

K16.273PPT (1)

Find the temperature of the first measurement: ( )

K401.001.4680K16.273

torr500torr734K16.2731

==

⎟⎟⎠

⎞⎜⎜⎝

⎛=T

Find the temperature of the third measurement: ( )

K400.591.4655K16.273

torr100torr65.146K16.2733

==

⎟⎟⎠

⎞⎜⎜⎝

⎛=T


1351

Assume a linear relationship between P/P3 and P3:

33

bPaPP

+=

where a is the pressure ratio for P3 = 0.

Substitute using the data from the first measurement:

( )torr500torr500torr734 ba +=

or ( )torr5004680.1 ba +=

Substitute using the data from the third measurement:

( )torr001torr001

torr65.461 ba +=

or ( )torr1004665.1 ba +=

Solve these equations simultaneously for a:

a = 1.46613

Substitute in equation (1) to obtain: ( ) K49.40046613.1K16.273 ==T

*74 •• Picture the Problem Because the O2 molecule resembles 2 spheres stuck together, which in cross section look something like two circles, we can estimate the radius of the molecule from the formula for the area of a circle. We can express the area, and hence the radius, of the circle in terms of the mean free path and the number density of the molecules and use the ideal-gas law to express the number density. Express the area of two circles of diameter d that touch each other: 24

222 ddA ππ

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

Solve for d to obtain:

πAd 2

= (1)

Relate the mean free path of the molecules to their number density and cross-sectional area:

Anv

1=λ

Solve for A to obtain: λvn

A 1=


λπ vnd 2=

Chapter 17

1352

Use the ideal-gas law to relate the number density of the O2 molecules to their temperature and pressure:

NkTPV = or kTP

VNnv ==


λπPkTd 2

=


( )( )( )( )

nm0.606m1006.6

m101.7Pa1001.1K300J/K10381.12

10

85

23

=×=

×××

=

−

−

−

πd

75 •• Picture the Problem We can use its definition to express the mean free path of the molecules and the ideal-gas law to obtain an expression for the number density of the hydrogen gas molecules. (a) Relate the mean free path of the molecules to their number density and cross-sectional area: Anv

1=λ

Use the ideal-gas law to relate the number density of the H2 molecules to their temperature and pressure:

NkTPV = or kTP

VNnv ==

Express the effective cross-sectional area of a H2 molecule:

241 dA π=

Substitute for nv and A to obtain: 2

4dP

kTπ

λ =

Substitute numerical values and evaluate λ:

( )( )( ) ( )

m1004.2

m106.1N/m1001.1K300J/K10381.14

6

21025

23

−

−

−

×=

××

×=π

λ

(b) Relate the available volume per molecule to the number density nv: P

kTnN

V

v

==1

Substitute numerical values and evaluate V/N:

( )( )

326

25

23

m1010.4

N/m1001.1K300J/K10381.1

−

−

×=

××

=NV

Express the volume of a spherical molecule:

3613

34 drV ππ ==


1353

Solve for d: 3

6πVd =


( ) nm28.4m1010.463

326

=×

=−

πd

1000. offactor aely approximatbylarger ispath freemean The

76 ••• Picture the Problem Let A be the cross-sectional area of the cylinder. We can use the ideal-gas law to find the height of the piston under equilibrium conditions. In (b), we can apply Newton’s 2nd law and the ideal-gas law for a fixed amount of gas to the show that, for small displacements from its equilibrium position, the piston executes simple harmonic motion.

(a) Express the pressure inside the cylinder: A

MgPP += atmin

Apply the ideal-gas law to obtain a second expression for the pressure of the gas in the cylinder:

hAnRT

VnRTP ==in (1)

Equate these two expressions: hA

nRTA

MgP =+atm

Solve for h to obtain: ( )

atm

atm

atm

atm

1

m4.2

m4.2

APMg

MgAPAP

MgAPnRTh

+=

+=

+=

At STP, 0.1 mol of gas occupies 2.24 L. Therefore:

( ) 33 m1024.2m4.2 −×=A

and A = 9.33×10−4 m2


( )( )m096.2

kPa101.3m109.333m/s9.81kg1.41

m2.4

24

2

=

×+

=

−

h

Chapter 17

1354

(b) Relate the frequency of vibration of the piston to its mass and a ″stiffness″ constant:

Mkf

π21

= (2)

where M is the mass of the piston and k is a constant of proportionality.

Letting y be the displacement from equilibrium, apply ∑ = yy maF to

the piston in its equilibrium position:

0atmin =−− APmgAP

For a small displacement y above equilibrium:

ymaAPmg'AP =−− atmin

or ymaAP'AP =− inin (3)

Using the ideal-gas law for a fixed amount of gas and constant temperature, relate inin to P'P :

VP'V'P inin =

or ( ) VPAyV'P inin =+

Solve for 'Pin : AyV

VP'P+

= inin

and

hyAP

AyAhAhAP'AP

+=

+=

1

1ininin


ymaAPhyAP =−⎟⎠⎞

⎜⎝⎛ +

−

in

1

in 1

or, for y << h,

ymaAPhyAP ≈−⎟⎠⎞

⎜⎝⎛ − inin 1 (4)

Simplify equation (4):

ymahyAP ≈− in


ymahyA

AhnRT

≈⎟⎠⎞

⎜⎝⎛−

or

ymayh

nRT≈⎟

⎠⎞

⎜⎝⎛− 2


1355

Solve for ay: ymhnRTay 2−=

or

ymkay −= , the condition for SHM

where 2mhnRT

mk=


221

mhnRTf

π=


( )( )( )( )( )

Hz01.1m2.096kg1.4

K300KJ/mol8.314mol0.121

2 =⋅

=π

f

*77 •••

Picture the Problem We can show that ( ) ( )∫ =V

xIdvvf0

, where f(v) is the Maxwell-

Boltzmann distribution function, ,kTmVx 22= and I(x) is the integral whose values are tabulated in the problem statement. Then, we can use this table to find the value of x corresponding to the fraction of the gas molecules with speeds less than v by evaluating I(x). (a) The Maxwell-Boltzmann speed distribution f(x) is given by:

( ) kTmvevkTmvf 22

232

24 −⎟

⎠⎞

⎜⎝⎛=

π

which means that the fraction of particles with speeds between v and v + dv is ( ) .dvvf

Express the fraction F(V) of particles with speeds less than V = 400 m/s:

( ) ( )

∫

∫

−⎟⎠⎞

⎜⎝⎛=

=

VkTmv

V

dvevkTm

dvvfVF

0

2223

0

2

24π

Change integration variables by letting kTmvz 2= so we can use the table of values to evaluate the integral. Then:

zmkTv 2

= ⇒ dzmkTdv 2

=

Chapter 17

1356

Substitute in the integrand of F(V) to obtain:

dzezmkT

dzmkTe

mkTzdvev

z

zkTmv

2

22

223

21222

2

22

−

−−

⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛=

Transform the integration limits to correspond to the new integration variable :2kTmvz =

When v = 0, z = 0, and when v = V, kTmVz 2=

The new lower integration limit is 0. Evaluate kTmVz 2= to find the upper limit:

( ) ( )( )( )( ) 06.1

K273K/J101.3812kg10661.1u32m/s400 23

27

=×

×= −

−

z

Evaluate F(400 m/s) to obtain:

( ) ( )

( )06.1

42

4m/s40006.1

0

2m/s400

0

2223m/s400

0

22

I

dzezdvevkTmdvvfF zkTmv

=

=⎟⎠⎞

⎜⎝⎛== ∫∫∫ −−

ππ

where ( ) ∫ −=x

z dzezxI0

2 24π

Letting r represent the fraction of the molecules with speeds less than 400 m/s, interpolate from the table to obtain:

15.1438.0788.0

106.1438.0

−−

=−

−r

and %0.48=r

(b) Express the fraction r of the molecules with speeds between V1 = 190 m/s and V2 = 565 m/s:

( ) ( ) ( ) ( )1212 xIxIVFVFr −=−= where

kTmVx 211 = and kTmVx 222 =

Evaluate x1 and x2 to obtain:

( ) ( )( )( )( ) 504.0

K273K/J101.3812kg10661.1u32m/s190 23

27

1 =×

×= −

−

x

and

( ) ( )( )( )( ) 50.1

K273K/J101.3812kg10661.1u32m/s565 23

27

2 =×

×= −

−

x

Substitute to obtain: ( ) ( )504.050.1 IIr −= (1)


1357

Using the table, evaluate I(1.50): ( ) 788.050.1 =I

Letting r represent the fraction of the molecules with speeds less than 190 m/s, interpolate from the table to obtain:

5.06.0081.0132.0

5.0504.0081.0

−−

=−

−r

and 083.0=r

Substitute in equation (1) to obtain: %5.70083.0788.0 =−=r

Chapter 17

1358

1359

Chapter 18 Heat and the First Law of Thermodynamics Conceptual Problems 1 • Picture the Problem We can use the relationship TmcQ ∆= to relate the temperature

changes of bodies A and B to their masses, specific heats, and the amount of heat supplied to each. Express the change in temperature of body A in terms of its mass, specific heat, and the amount of heat supplied to it:

AAA cm

QT =∆

Express the change in temperature of body B in terms of its mass, specific heat, and the amount of heat supplied to it:

BBB cm

QT =∆


BB

AA

A

B

cmcm

TT

=∆∆


( )( ) 422

BB

BB

A

B ==∆∆

cmcm

TT

or

AB 4 TT ∆=∆

*2 • Picture the Problem We can use the relationship TmcQ ∆= to relate the temperature

changes of bodies A and B to their masses, specific heats, and the amount of heat supplied to each. Relate the temperature change of block A to its specific heat and mass:

AAA cM

QT =∆

Relate the temperature change of block B to its specific heat and mass:

BBB cM

QT =∆

Chapter 18

1360

Equate the temperature changes to obtain:

AABB

11cMcM

=

Solve for cA: B

A

BA c

MMc =

and correct. is )( b

3 • Picture the Problem We can use the relationship TmcQ ∆= to relate the amount of

energy absorbed by the aluminum and copper bodies to their masses, specific heats, and temperature changes. Express the energy absorbed by the aluminum object:

TcmQ ∆= AlAlAl

Express the energy absorbed by the copper object:

TcmQ ∆= CuCuCu


TcmTcm

QQ

∆∆

=AlAl

CuCu

Al

Cu

Because the object’s masses are the same and they experience the same change in temperature:

1Al

Cu

Al

Cu <=cc

QQ

or

AlCu QQ < and correct. is )( c

4 • Determine the Concept Some examples of systems in which internal energy is converted into mechanical energy are: a steam turbine, an internal combustion engine, and a person performing mechanical work, e.g., climbing a hill. *5 • Determine the Concept Yes, if the heat absorbed by the system is equal to the work done by the system. 6 • Determine the Concept According to the first law of thermodynamics, the change in the internal energy of the system is equal to the heat that enters the system plus the work done on the system. correct. is )( b

Heat and the First Law of Thermodynamics

1361

7 • Determine the Concept .oninint WQE +=∆ For an ideal gas, ∆Eint is a function of T

only. Because Won = 0 and Qin = 0 in a free expansion, ∆Eint = 0 and T is constant. For a real gas, Eint depends on the density of the gas because the molecules exert weak attractive forces on each other. In a free expansion, these forces reduce the average kinetic energy of the molecules and, consequently, the temperature. 8 • Determine the Concept Because the container is insulated, no energy is exchanged with the surroundings during the expansion of the gas. Neither is any work done on or by the gas during this process. Hence, the internal energy of the gas does not change and we can conclude that the equilibrium temperature will be the same as the initial temperature. Applying the ideal-gas law for a fixed amount of gas we see that the pressure at equilibrium must be half an atmosphere. correct. is )( c

9 • Determine the Concept The temperature of the gas increases. The average kinetic energy increases with increasing volume due to the repulsive interaction between the ions. *10 •• Determine the Concept The balloon that expands isothermally is larger when it reaches the surface. The balloon that expands adiabatically will be at a lower temperature than the one that expands isothermally. Because each balloon has the same number of gas molecules and are at the same pressure, the one with the higher temperature will be bigger. An analytical argument that leads to the same conclusion is shown below. Letting the subscript ″a″ denote the adiabatic process and the subscript ″i″ denote the isothermal process, express the equation of state for the adiabatic balloon:

γγaf,f00 VPVP = ⇒

γ1

f

00af, ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

For the isothermal balloon: if,f00 VPVP = ⇒ ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

f

00if, P

PVV

Divide the second of these equations by the first and simplify to obtain:

λ

γ

11

f

01

f

00

f

00

af,

if,−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

=PP

PPV

PPV

VV

Chapter 18

1362

Because P0/Pf > 1 and γ > 1: af,if, VV >

11 • Determine the Concept The work done along each of these paths equals the area under its curve. The area is greatest for the path A→B→C and least for the path A→D→C.

correct. is )( a

12 • Determine the Concept An adiabatic process is, by definition, one for which no heat enters or leaves the system. correct. is )( b

13 • (a) False. The heat capacity of a body is the heat needed to raise the temperature of the body by one degree. (b) False. The amount of heat added to a system when it goes from one state to another is path dependent. (c) False. The work done on a system when it goes from one state to another is path dependent. (d) True. (e) True. (f) True. (g) True. *14 • Determine the Concept For a constant-volume process, no work is done on or by the gas. Applying the first law of thermodynamics, we obtain Qin = ∆Eint. Because the temperature must change during such a process, we can conclude that ∆Eint ≠ 0 and hence Qin ≠ 0. correct. are )( and )( db

15 • Determine the Concept Because the temperature does not change during an isothermal process, the change in the internal energy of the gas is zero. Applying the first law of thermodynamics, we obtain Qin = −Won = Wby the system. Hence correct. is )(d


1363

16 •• Determine the Concept The melting point of propane at 1 atm pressure is 83 K. Hence, at this low temperature and high pressure, C3H8 is a solid. 17 •• Picture the Problem We can use the given dependence of the pressure on the volume and the ideal-gas law to show that if the volume decreases, so does the temperature. We’re given that: constant=VP

Because the gas is an ideal gas: ( ) nRTVVVPPV === constant

Solve for T: ( )

nRVT constant

=

decreases.re temperatuthe decreases, volume theif , ofroot square with the varies Because VT

*18 •• Determine the Concept At room temperature, most solids have a roughly constant heat capacity per mole of 6 cal/mol-K (Dulong-Petit law). Because 1 mole of lead is more massive than 1 mole of copper, the heat capacity of lead should be lower than the heat capacity of copper. This is, in fact, the case. 19 •• Determine the Concept The heat capacity of a substance is proportional to the number of degrees of freedom per molecule associated with the molecule. Because there are 6 degrees of freedom per molecule in a solid and only 3 per molecule (translational) for a monatomic liquid, you would expect the solid to have the higher heat capacity. Estimation and Approximation *20 •• Picture the Problem The heat capacity of lead is c = 128 J/kg⋅K. We’ll assume that all of the work done in lifting the bag through a vertical distance of 1 m goes into raising the temperature of the lead shot and use conservation of energy to relate the number of drops of the bag and the distance through which it is dropped to the heat capacity and change in temperature of the lead shot. (a) Use conservation of energy to relate the change in the potential energy of the lead shot to the change in its temperature:

TmcNmgh ∆= where N is the number of times the bag of shot is dropped.

Chapter 18

1364

Solve for ∆T to obtain: c

Nghmc

NmghT ==∆

Substitute numerical values and evaluate ∆T:

( )( ) K83.3KJ/kg128

m1m/s`81.950 2

=⋅

=∆T

(b)

increases. mass theas decreases which );L /L(L mass itsby dividedshot theof area surface

theas slost varieheat ofamount The mass. its toalproportion isheat gainsit at which rate the whilearea surface its toalproportion isshot leadby thelost isheat at which rate thebecause masslarger a use better to isIt

132 −=

21 •• Picture the Problem Assume that the water is initially at 30°C and that the cup contains 200 g of water. We can use the definition of power to express the required time to bring the water to a boil in terms of its mass, heat capacity, change in temperature, and the rate at which energy is supplied to the water by the microwave oven. Use the definition of power to relate the energy needed to warm the water to the elapsed time:

tTmc

tWP

∆∆

=∆∆

=

Solve for ∆t to obtain: P

Tmct ∆=∆


( )( )( ) min1.63s5.97W600

K330K733KkJ/kg18.4kg2.0==

−⋅=∆t , an elapsed time

that seems to be consistent with experience. 22 • Picture the Problem The adiabatic compression from an initial volume V1 to a final volume V2 between the isotherms at temperatures T1 and T2 is shown to the right. We’ll assume a room temperature of 300 K and apply the equation for a quasi-static adiabatic process with γair = 1.4 to solve for the ratio of the initial to the final volume of the air.

Express constant1 =−γTV in terms of the initial and final values of T and V:

122

111

−− = γγ VTVT


1365

Solve for V1/V2 to obtain: 1

1

1

2

2

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γ

TT

VV

Substitute numerical values and evaluate V1/V2:

69.3K300K506 14.1

1

2

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

VV

23 •• Picture the Problem We can use TmcQ ∆= p to express the specific heat of water during heating at constant pressure in terms of the required heat and the resulting change in temperature. Further, we can use the definition of the bulk modulus to express the work done by the water as it expands. Equating the work done by the water during its expansion and the heat gained during this process will allow us to solve for cp. Express the heat needed to raise the temperature of a mass m of a substance whose specific heat at constant pressure is cp by ∆T:

TmcQ ∆= p

Solve for cp to obtain: Tm

Qc∆

=p

Use the definition of the bulk modulus to express the work done by the water as it expands:

VPV

VVPB

∆∆

=∆∆

=

or VBPVW ∆=∆=

Assuming that the work done by the water in expanding equals the heat gained during the process, substitute to obtain:

TmVBc∆∆

=p

Using the definition of the coefficient of volume expansion, express ∆V (see Chapter 20, Section 1):

TVV ∆=∆ β

Substitute to obtain: m

VBTm

TVBc ββ=

∆∆

=p

Chapter 18

1366

Use the data given in the problem statement to find the average volume of 1 kg of water as it warms from 4°C to 100°C:

33

33

m1002.12

g/cm9584.0g/cm1.0000kg1

−×=

+=

=ρmV

Substitute numerical values and evaluate cp:

( )( )( ) KJ/kg2.42kg1

m1002.1K10207.0N/m102 33-1328

p ⋅=×××

=−−

c

Express the ratio of cp to cwater: 2

water

p 1001.1KJ/kg4184KJ/kg2.42 −×=⋅⋅

=c

c

or ( ) waterp %01.1 cc =

*24 •• Picture the Problem We can apply the condition for the validity of the equipartition theorem, i.e., that the spacing of the energy levels be large compared to kT, to find the critical temperature Tc: Express the failure condition for the equipartition theorem:

eV15.0c ≈kT

Solve for Tc: k

T eV15.0c =

Substitute numerical values and evaluate Tc: K1740

J/K101.381eV

J101.602eV0.1523

19

c =×

××

= −

−

T

Heat Capacity; Specific Heat; Latent Heat *25 • Picture the Problem We can use the conversion factor 1 cal = 4.184 J to convert 2500 kcal into joules and the definition of power to find the average output if the consumed energy is dissipated over 24 h. (a) Convert 2500 kcal to joules:

MJ5.10cal

J4.184kcal2500kcal2500

=

×=


1367

(b) Use the definition of average power to obtain: W121

hs3600h24

J 1005.1 7

av =×

×=

∆∆

=tEP

Remarks: Note that this average power output is essentially that of a widely used light bulb. 26 • Picture the Problem We can use the relationship TmcQ ∆= to calculate the amount of

heat given off by the concrete as it cools from 25 to 20°C. Relate the heat given off by the concrete to its mass, specific heat, and change in temperature:

TmcQ ∆=

Substitute numerical values and evaluate Q:

( )( )( )MJ500

K293K298KkJ/kg1kg105

=

−⋅=Q

27 • Picture the Problem We can find the amount of heat that must be supplied by adding the heat required to warm the ice from −10°C to 0°C, the heat required to melt the ice, and the heat required to warm the water formed from the ice to 40°C. Express the total heat required: waterwarmicemelticewarm QQQQ ++=

Substitute for each term to obtain: ( )waterwaterficeice

waterwaterficeice

TcLTcmTmcmLTmcQ

∆++∆=∆++∆=


( ) ( )( )[( )( )]

kcal48.7

K273K313Kkcal/kg1kcal/kg79.7K263K273Kkcal/kg49.0kg0.06

=

−⋅++−⋅=Q

28 •• Picture the Problem We can find the amount of heat that must be removed by adding the heat that must be removed to cool the steam from 150°C to 100°C, the heat that must be removed to condense the steam to water, the heat that must be removed to cool the water from 100°C to 0°C, and the heat that must be removed to freeze the water.

Chapter 18

1368

Express the total heat that must be removed:

waterfreezewatercool

steamcondensesteamcool

QQ

QQQ

++

+=

Substitute for each term to obtain:

( )fwaterwatervsteamsteam

fwaterwater

vsteamsteam

LTcLTcmmLTmc

mLTmcQ

+∆++∆=+∆+

+∆=


( ) ( )( )[( )( ) ]

kcal4.74

kJ4.184kcal1kJ2.311

kJ/kg5.333K273K373KkJ/kg18.4MJ/kg26.2K373K423KkJ/kg01.2kg0.1

=

×=

+−⋅++−⋅=Q

29 •• Picture the Problem We can find the amount of nitrogen vaporized by equating the heat gained by the liquid nitrogen and the heat lost by the piece of aluminum. Express the heat gained by the liquid nitrogen as it cools the piece of aluminum:

vNNN LmQ =

Express the heat lost by the piece of aluminum as it cools:

AlAlAlAl TcmQ ∆=

Equate these two expressions and solve for mN:

AlAlAlvNN TcmLm ∆=

and

vN

AlAlAlN L

Tcmm ∆=

Substitute numerical values and evaluate mN:

( )( )( ) mg48.8kg1088.4kJ/kg199

K77K293KJ/kg0.90kg0.05 5N =×=

−⋅= −m


1369

30 •• Picture the Problem Because the heat lost by the lead as it cools is gained by the block of ice (we’re assuming no heat is lost to the surroundings), we can apply the conservation of energy to determine how much ice melts. Apply the conservation of energy to this process:

0=∆Q

or ( ) 0wf,wPbPbPbf,Pb =+∆+− LmTcLm

Solve for mw:

wf,

PbPbPbf,Pbw L

TcLmm

⎟⎠⎞⎜

⎝⎛ ∆+

=

Substitute numerical values and evaluate mw:

( ) ( )( )( ) g8.99kJ/kg333.5

K273K600KkJ/kg0.128kJ/kg7.24kg5.0w =

−⋅+=m

*31 •• Picture the Problem The temperature of the bullet immediately after coming to rest in the block is the sum of its pre-collision temperature and the change in its temperature as a result of being brought to a stop in the block. We can equate the heat gained by the bullet and half its pre-collision kinetic energy to find the change in its temperature. Express the temperature of the bullet immediately after coming to rest in terms of its initial temperature and the change in its temperature as a result of being stopped in the block:

TTTT∆+=

∆+=K293

i

Relate the heat absorbed by the bullet as it comes to rest to its kinetic energy before the collision:

KQ 21=

Substitute for Q and K to obtain: ( )2Pb2

121

PbPb vmTcm =∆

Solve for ∆T:

Pb

2

4cvT =∆

Chapter 18

1370


Pb

2

4K293

cvT +=


( )( )

C365K638

KkJ/kg0.1284m/s420K293

2

°==

⋅+=T

32 •• Picture the Problem We can find the heat available to warm the brake drums from the initial kinetic energy of the car and the mass of steel contained in the brake drums from Q = msteelcsteel∆T. Express msteel in terms of Q:

TcQm∆

=steel

steel

Find the heat available to warm the brake drums from the initial kinetic energy of the car:

2car2

1 vmKQ ==

Substitute for Q to obtain: Tcvmm∆

=steel

2car2

1

steel

Substitute numerical values and evaluate msteel: ( )

( )

kg6.26

K120kcal

kJ4.186Kkg

kcal0.112

s3600h1

hkm80kg1400

2

steel

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

⋅

⎟⎟⎠

⎞⎜⎜⎝

⎛×

=m

Calorimetry 33 • Picture the Problem Let the system consist of the piece of lead, calorimeter, and water. During this process the water will gain energy at the expense of the piece of lead. We can set the heat out of the lead equal to the heat into the water and solve for the final temperature of the lead and water. Apply conservation of energy to the system to obtain:

0=∆Q or outin QQ =


1371

Express the heat lost by the lead in terms of its specific heat and temperature change:

PbPbPbout TcmQ ∆=

Express the heat absorbed by the water in terms of its specific heat and temperature change:

wwwin TcmQ ∆=


PbPbPbwww TcmTcm ∆=∆

Substitute numerical values:

( )( )( ) ( )( )( )ff K363KkJ/kg128.0kg0.2K293KkJ/kg18.4kg0.5 tt −⋅=−⋅

Solve for tf to obtain: C8.20K8.293f °==t

*34 • Picture the Problem During this process the water and the container will gain energy at the expense of the piece of metal. We can set the heat out of the metal equal to the heat into the water and the container and solve for the specific heat of the metal. Apply conservation of energy to the system to obtain:

0=∆Q or lostgained QQ =

Express the heat lost by the metal in terms of its specific heat and temperature change:

metalmetalmetallost TcmQ ∆=

Express the heat gained by the water and the container in terms of their specific heats and temperature change:

wmetalcontainerwwwgained TcmTcmQ ∆+∆=


metalmetalmetalwmetalcontainerwww TcmTcmTcm ∆=∆+∆ Substitute numerical values:

( )( )( ) ( )( )( )( ) metal

metal

K4.294K373kg0.1K293K4.294kg0.2K293K4.294KkJ/kg18.4kg0.5

cc

−=−+−⋅

Chapter 18

1372

Solve for cmetal: KkJ/kg386.0metal ⋅=c

35 •• Picture the Problem We can use TmcQ ∆= to express the mass m of water that can be heated through a temperature interval ∆T by an amount of heat energy Q. We can then find the amount of heat energy expended by Armstrong from the definition of power. Express the amount of heat energy Q required to raise the temperature of a mass m of water by ∆T:

TmcQ ∆=

Solve for m to obtain:

TcQm∆

=

Use the definition of power to relate the heat energy expended by Armstrong to the rate at which he expended the energy:

tQP∆

= ⇒ tPQ ∆=

Substitute to obtain: Tc

tPm∆∆

=


( )( )( )( )( )( )

kg453

K792K373KkJ/kg184.4d20h/d5s/h3600J/s400

=

−⋅=m

36 •• Picture the Problem During this process the ice and the water formed from the melted ice will gain energy at the expense of the glass tumbler and the water in it. We can set the heat out of the tumbler and the water that is initially at 24°C equal to the heat into the ice and ice water and solve for the final temperature of the drink. Apply conservation of energy to the system to obtain:


Express the heat lost by the tumbler and the water in it in terms of their specific heats and common temperature change:

TcmTcmQ ∆+∆= waterwaterglassglasslost

Express the heat gained by the ice and the melted ice in terms of their specific heats and temperature

watericewaterwaterice

ficeiceiceicegained

Tcm

LmTcmQ

∆+

+∆=


1373

changes: Substitute to obtain:

TcmTcmTcmLmTcm ∆+∆=∆++∆ waterwaterglassglasswatericewaterwatericeficeiceiceice


( )( )( ) ( )( )( )( ) ( )( )( )

( )( )( )f

ff

K297Kkcal/kg1kg0.2K297Kkcal/kg2.0kg0.025Kkcal/kg1kg0.03

kcal/kg7.79kg0.03K270K273Kkcal/kg49.0kg0.03

ttt

−⋅+−⋅=⋅+

+−⋅

Solve for tf: C6.10K6.283f °==t

37 •• Picture the Problem Because we can not tell, without performing a couple of calculations, whether there is enough heat available in the 500 g of water to melt all of the ice, we’ll need to resolve this question first. (a) Determine the heat required to melt 200 g of ice: ( )( )

kcal15.94kcal/kg79.7kg0.2

ficeicemelt

==

= LmQ

Determine the heat available from 500 g of water: ( )( )

( )kcal10

K273K932Kkcal/kg1kg0.5

waterwaterwaterwater

=−×

⋅=∆= TcmQ

Because Qwater < Qmelt ice: C.0 is re temperatufinal The °

(b) Equate the energy available from the water Qwater to miceLf and solve for mice:

f

waterice L

Qm =

Substitute numerical values and evaluate mice:

g125kcal/kg79.7kcal10

ice ==m

Chapter 18

1374

38 •• Picture the Problem Because the bucket contains a mixture of ice and water initially, we know that its temperature must be 0°C. We can equate the heat gained by the mixture of ice and water and the heat lost by the block of copper and solve for the amount of ice initially in the bucket. Apply conservation of energy to the system to obtain:


Express the heat lost by the block of copper:

CuCuCulost TcmQ ∆=

Express the heat gained by the ice and the melted ice:

watericewaterwatericeficegained TcmLmQ ∆+=


0CuCuCu

watericewaterwatericefice

=∆−

∆+

Tcm

TcmLm

Solve for mice:

f

watericewaterwatericeCuCuCuice L

TcmTcmm

∆−∆=

Substitute numerical values and evaluate mice:

( )( )( )

( )( )( )

g171

kcal/kg79.7K273K281Kkcal/kg1kg2.1

kcal/kg79.7K281K353Kkcal/kg0923.0kg5.3

ice

=

−⋅−

−⋅=m

39 •• Picture the Problem During this process the ice and the water formed from the melted ice will gain energy at the expense of the condensing steam and the water from the condensed steam. We can equate these quantities and solve for the final temperature of the system. (a) Apply conservation of energy to the system to obtain:


Express the heat required to melt the ice and raise the temperature of the

waterwaterwatericeficegained TcmLmQ ∆+=


1375

ice water: Express the heat available from 20 g of steam and the cooling water formed from the condensed steam:

waterwatersteamvsteamlost TcmLmQ ∆+=


waterwatersteamvsteamwaterwaterwatericefice TcmLmTcmLm ∆+=∆+


( )( ) ( )( )( )( )( ) ( )( )( )f

f

K733Kkcal/kg1kg0.02kcal/kg540kg0.02K273Kkcal/kg1kg15.0kcal/kg79.7kg0.15

tt

−⋅+=−⋅+

Solve for tf:

C4.94K94.277f °==t

(b) left. is ice no C,0an greater th is e temperturfinal theBecause °

40 •• Picture the Problem During this process the ice will gain heat and the water will lose heat. We can do a preliminary calculation to determine whether there is enough heat available to melt all of the ice and, if there is, equate the heat the heat lost by the water to the heat gained by the ice and resulting ice water as the system achieves thermal equilibrium. Apply conservation of energy to the system to obtain:


Find the heat available to melt the ice: ( )( )( )kcal30

K273K303Kkcal/kg1kg1

waterwaterwateravail

=−×

⋅=∆= TcmQ

Find the heat required to melt all of the ice: ( )( )

kcal3.985kcal/kg79.7kg0.05

ficeicemelt

==

= LmQ

Because Qavail > Qmelt ice, we know waterwaterwaterlost TcmQ ∆=

Chapter 18

1376

that the final temperature will be greater than 273 K and we can express Qlost in terms of the change in temperature of the water: Express Qgained:

watericewaterwatericeficegained TcmLmQ ∆+=

Equate the heat gained and the heat lost to obtain:

waterwaterwaterwatericewaterwatericefice TcmTcmLm ∆=∆+


( )( ) ( )( )( )( )( )( )f

f

K303Kkcal/kg1kg1K273Kkcal/kg1kg05.0kcal/kg7.79kg05.0

TT

−⋅=−⋅+

Solving for Tf yields: C24.8K297.8f °==T

Find the heat required to melt 500 g of ice:

( )( )kcal39.85

kcal/kg79.7kg0.5ficeicemelt

==

= LmQ

C.0 be willre temperatufinal theavailable,heat an thegreater th is ice of g 500melt torequiredheat theBecause

°

*41 •• Picture the Problem Assume that the calorimeter is in thermal equilibrium with the water it contains. During this process the ice will gain heat in warming to 0°C and melting, as will the water formed from the melted ice. The water in the calorimeter and the calorimeter will lose heat. We can do a preliminary calculation to determine whether there is enough heat available to melt all of the ice and, if there is, equate the heat the heat lost by the water to the heat gained by the ice and resulting ice water as the system achieves thermal equilibrium. Find the heat available to melt the ice:

( )( )[ ( )( )]( )kJ40.45

K 273K932KkJ/kg9.0kg2.0KkJ/kg18.4kg5.0watercalcalwaterwaterwateravail

=−⋅+⋅=

∆+∆= TcmTcmQ


1377

Find the heat required to melt all of the ice:

( )( )( ) ( )( )kJ35.37

kJ/kg5.333kg0.1K253K372KkJ/kg2kg0.1ficeiceiceiceicemelt

=+−⋅=

+∆= LmTcmQ

(a) Because Qavail > Qmelt ice, we know that the final temperature will be greater than 0°C. Apply the conservation of energy to the system to obtain:


Express Qlost in terms of the final temperature of the system:

( ) rcalorimetewatercalcalwaterwaterlost +∆+= TcmcmQ

Express Qgained in terms of the final temperature of the system:

ficeiceiceicegained LmTcmQ +∆=


( ) rcalorimetewatercalcalwaterwaterwatericewaterwatericeficeiceiceice +∆+=∆++∆ TcmcmTcmLmTcm


( )( )( )( )( )[ ( )( )]( )f

f

K932KkJ/kg0.9kg2.0KkJ/kg4.18kg5.0K273KkJ/kg18.4kg1.0kJ35.37

tt

−⋅+⋅=−⋅+

Solving for tf yields: C99.2K276f °==t

(b) Find the heat required to raise 200 g of ice to 0°C:

( )( )( ) kJ00.8K253K273KkJ/kg2kg0.2iceiceiceicewarm =−⋅=∆= TcmQ

Noting that there are now 600 g of water in the calorimeter, find the heat available from cooling the calorimeter and water from 3°C to 0°C:

( )( )[ ( )( )]( )kJ064.8

K273K293KkJ/kg9.0kg2.0KkJ/kg18.4kg6.0watercalcalwaterwaterwateravail

=−⋅+⋅=

∆+∆= TcmTcmQ

Chapter 18

1378

Express the amount of ice that will melt in terms of the difference between the heat available and the heat required to warm the ice:

f

ice warmavailicemelted L

QQm −=

Substitute numerical values and evaluate mmelted ice: g0.1919

kJ/kg333.5kJ8kJ8.064

icemelted

=

−=m

Find the ice remaining in the system: g199.8

g0.1919g002iceremaining

=

−=m

(c) same. the

be ldanswer wou thesame, theare conditions final and initial theBecause

42 •• Picture the Problem Let the subscript B denote the block, w1 the water initially in the calorimeter, and w2 the 120 mL of water that is added to the calorimeter vessel. We can equate the heat gained by the calorimeter and its initial contents to the heat lost by the warm water and solve this equation for the specific heat of the block. Apply conservation of energy to the system to obtain:


Express the heat gained by the block, the calorimeter, and the water initially in the calorimeter:

( ) Tcmcmcm

Tcm

TcmTcmQ

∆++=

∆+

∆+∆=

11

111

wwCuCuBB

www

CuCuCuBBBgained

because the temperature changes are the same for the block, calorimeter, and the water that is initially at 20°C.

Express the heat lost by the water that is added to the calorimeter:

222 wwwlost TcmQ ∆=


( )22211 wwwwwCuCuBB TcmTcmcmcm ∆=∆++


1379

Substitute numerical values to obtain: ( ) ( )[ ( ) ( )( )]( )

( )( )( )K327K353KkJ/kg4.18kg10120

K293K327KkJ/kg4.18kg0.06KkJ/kg0.386kg025.0kg1.03

B

−⋅×=

−⋅+⋅+−

c

Solve for cB to obtain:

Kcal/g294.0

KkJ/kg23.1B

⋅=

⋅=c

43 •• Picture the Problem We can find the temperature t by equating the heat gained by the warming water and calorimeter, and vaporization of some of the water. Apply conservation of energy to the system to obtain:


Express the heat gained by the warming and vaporizing water:

wcalcal

wwwwf,vaporizedw,gained

Tcm

TcmLmQ

∆+

∆+=

Express the heat lost by the 100-g piece of copper as it cools:

CuCuCulost TcmQ ∆=


CuCuCuwcalcalwwwwf,vaporizedw, TcmTcmTcmLm ∆=∆+∆+

Substitute numerical values: ( )( ) ( )( )( )

( )( )( ) ( )( )( )K311Kcal/g0.0923g100K892K113Kcal/g0923.0g150K892K113Kcal/g1g200cal/g540g1.2

−⋅=−⋅+−⋅+

t

Solve for t to obtain: C618K891 °==t

44 ••

Picture the Problem We can find the final temperature of the system by equating the heat gained by the calorimeter and the water in it to the heat lost by the cooling aluminum shot. In (b) we’ll proceed as in (a) but with the initial and final temperatures adjusted to minimize heat transfer between the system and its surroundings.

Chapter 18

1380

Apply conservation of energy to the system to obtain:


(a) Express the heat gained by the warming water and the calorimeter:

wAlcalwwwgained TcmTcmQ ∆+∆=

Express the heat lost by the aluminum shot as it cools:

AlAlshotlost TcmQ ∆=


( ) AlAlshotwAlcalww TcmTcmcm ∆=∆+


( )( )[ ( )( )]( )( )( )( )f

f

K373Kcal/g0.215g300K392Kcal/g0923.0g200Kcal/g1g500

tt

−⋅=−⋅+⋅

Solve for tf to obtain: C9.28K9.301f °==t

(b) Let the initial and final temperatures of the calorimeter and its contents be:

ti = 20°C – t0 (1) and tf = 20°C + t0 where ti and tf are the temperatures above and below room temperature and t0 is the amount ti and tf must be below and above room temperature respectively.

Express and the heat gained by the water and calorimeter:

( ) wAlcalww

wAlcalwwwin

TcmcmTcmTcmQ

∆+=∆+∆=

Express the heat lost by the aluminum shot as it cools:

AlAlshotout TcmQ ∆=

Equate Qin and Qout to obtain:

( ) AlAlshotwAlcalww TcmTcmcm ∆=∆+


( )( )[ ( )( )]( )( )( )( )0

00

K293K373Kcal/g0.215g300K293K293Kcal/g215.0g200Kcal/g1g500

ttt

−−⋅=+−+⋅+⋅

Solve for and evaluate t0: C49.4K49.2770 °==t


1381

Substitute in equation (1) to obtain: C5.15C49.4C20i °=°−°=t

First Law of Thermodynamics 45 • Picture the Problem We can apply the first law of thermodynamics to find the change in internal energy of the gas during this process. Apply the first law of thermodynamics to express the change in internal energy of the gas in terms of the heat added to the system and the work done on the gas:

oninint WQE +=∆

The work done by the gas equals the negative of the work done on the gas. Substitute numerical values and evaluate ∆Eint:

kJ2.21

J300cal

J4.184cal006int

=

−×=∆E

*46 • Picture the Problem We can apply the first law of thermodynamics to find the change in internal energy of the gas during this process. Apply the first law of thermodynamics to express the change in internal energy of the gas in terms of the heat added to the system and the work done on the gas:

oninint WQE +=∆

The work done by the gas is the negative of the work done on the gas. Substitute numerical values and evaluate ∆Eint:

kJ748

kJ800cal

J4.184kcal004int

=

−×=∆E

47 • Picture the Problem We can use the first law of thermodynamics to relate the change in the bullet’s internal energy to its pre-collision kinetic energy.

Chapter 18

1382

Using the first law of thermodynamics, relate the change in the internal energy of the bullet to the work done on it by the block of wood:

oninint WQE +=∆

or, because Qin = 0, ( )iKKKWE −−=∆==∆ fonint

Substitute for ∆Eint, Kf, and Ki to obtain:

( ) ( ) 2212

21

ifPb 0 mvmvttmc =−−=−

Solve for tf:

Pb

2

if 2cvtt +=

Substitute numerical values and evaluate tf:

( )( )

C176K449

KkJ/kg128.02m/s200K293

2

f

°==

⋅+=t

48 • Picture the Problem What is described above is clearly a limiting case because, as the water falls, it will, for example, collide with rocks and experience air drag; resulting in some of its initial potential energy being converted into internal energy. In this limiting case we can use the first law of thermodynamics to relate the change in the gravitational potential energy (take Ug = 0 at the bottom of the waterfalls) to the change in internal energy of the water and solve for the increase in temperature. (a) Using the first law of thermodynamics and noting that, because the gravitational force is conservative, Won = −∆U, relate the change in the internal energy of the water to the work done on it by gravity:

oninint WQE +=∆

or, because Qin = 0, ( )ifonint UUUWE −−=∆−==∆

Substitute for ∆Eint, Uf, and Ui to obtain:

( ) hmghmgTmc ∆=∆−−=∆ 0w

Solve for ∆T: wc

hgT ∆=∆

Substitute numerical values and evaluate ∆T:

( )( ) K117.0KkJ/kg4.18m50m/s9.81 2

=⋅

=∆T


1383

(b) Proceed as in (a) with ∆h = 740 m:

( )( ) K74.1KkJ/kg4.18

m740m/s9.81 2

=⋅

=∆T

49 • Picture the Problem We can apply the first law of thermodynamics to find the change in internal energy of the gas during this process. Apply the first law of thermodynamics to express the change in internal energy of the gas in terms of the heat added to the system and the work done on the gas:

oninint WQE +=∆

The work done by the gas is the negative of the work done on the gas. Substitute numerical values and evaluate ∆Eint:

J7.53J03cal

J4.184cal20int =−×=∆E

50 •• Picture the Problem We can use the definition of kinetic energy to express the speed of the bullet upon impact in terms of its kinetic energy. The heat absorbed by the bullet is the sum of the heat required to warm to bullet from 202 K to its melting temperature of 600 K and the heat required to melt it. We can use the first law of thermodynamics to relate the impact speed of the bullet to the change in its internal energy. Using the first law of thermodynamics, relate the change in the internal energy of the bullet to the work done on it by the target:

oninint WQE +=∆

or, because Qin = 0, ( )iKKKWE −−=∆==∆ fonint

Substitute for ∆Eint, Kf, and Ki to obtain:

( ) 2212

21

Pbf,PbPb 0 mvmvmLTmc =−−=+∆

or ( ) 2

21

Pbf,iMPPb mvmLTTmc =+−


( )[ ]Pbf,iMPPb2 LTTcv +−=


( )( ) m/s354kJ/kg24.7K303K600KkJ/kg0.1282 =+−⋅=v

Chapter 18

1384

*51 •• Picture the Problem We can find the rate at which heat is generated when you rub your hands together using the definition of power and the rubbing time to produce a 5°C increase in temperature from ( ) tdtdQQ ∆=∆ and

Q = mc∆T. (a) Express the rate at which heat is generated as a function of the friction force and the average speed of your hands:

vFvfPdtdQ

nk µ===

Substitute numerical values and evaluate dQ/dt:

( )( ) W6.13m/s0.35N35.50 ==dtdQ

(b) Relate the heat required to raise the temperature of your hands 5 K to the rate at which it is being generated:

TmctdtdQQ ∆=∆=∆

Solve for ∆t: dtdQTmct ∆

=∆


( )( )( )

min0.19s60

min1s1143

W6.13K5KkJ/kg4kg0.35

=×=

⋅=∆t

Work and the PV Diagram for a Gas 52 • Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isobaric expansion. We can then use the first law of thermodynamics to find the heat added to the system during this process.


1385

(a) The path from the initial state (1) to the final state (2) is shown on the PV diagram.

The work done by the gas equals the area under the shaded curve:

( )( ) J608L

m10L2atm

kPa101.3atm3L2atm333

gasby =⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎠⎞

⎜⎝⎛ ×==∆=

−

VPW

(b) The work done by the gas is the negative of the work done on the gas. Apply the first law of thermodynamics to the system to obtain:

( ) ( )( ) gasby int,1int,2

gasby int,1int,2

onintin

WEE

WEEWEQ

+−=

−−−=−∆=

Substitute numerical values and evaluate Qin:

( ) kJ1.06J608J456J912in =+−=Q

53 • Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isobaric expansion. We can then use the first law of thermodynamics to find the heat added to the system during this process (a) The path from the initial state (1) to the final state (2) is shown on the PV diagram.

Chapter 18

1386

The work done by the gas equals the area under the curve:

( )( ) J054L

m10L2atm

kPa101.3atm2L2atm233

gasby =⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎠⎞

⎜⎝⎛ ×==∆=

−

VPW



gasby int,1int,2

onintin

WEE

WEEWEQ

+−=

−−−=−∆=


( ) J618J054J456J912in =+−=Q

*54 •• Picture the Problem We can find the work done by the gas during this process from the area under the curve. Because no work is done along the constant volume (vertical) part of the path, the work done by the gas is done during its isothermal expansion. We can then use the first law of thermodynamics to find the heat added to the system during this process. (a) The path from the initial state (1) to the final state (2) is shown on the PV diagram.


[ ]

3ln

ln

1

L3L111

L3

L111

L3

L11gasby

2

1

VP

VVPVdVVP

VdVnRTdVPW

V

V

=

==

==

∫

∫∫

Substitute numerical values and evaluate Wby gas:


1387

J3343lnL

m10L1atm

kPa101.3atm333

gasby =⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎠⎞

⎜⎝⎛ ×=

−

W



gasby int,1int,2

onintin

WEE

WEEWEQ

+−=

−−−=−∆=


( ) J790J334J456J912in =+−=Q

55 •• Picture the Problem We can find the work done by the gas during this process from the area under the curve. We can then use the first law of thermodynamics to find the heat added to the system during this process. (a) The path from the initial state (1) to the final state (2) is shown on the PV diagram:


( )( )

J507LatmJ101.3Latm5

L2atm2atm321

trapezoidgasby

=⋅

×⋅=

+== AW



gasby int,1int,2

onintin

WEE

WEEWEQ

+−=

−−−=−∆=


( ) J639J507J456J912in =+−=Q

Chapter 18

1388

Remarks: You could use the linearity of the path connecting the initial and final states and the coordinates of the endpoints to express P as a function of V. You could then integrate this function between 1 and 3 L to find the work done by the gas as it goes from its initial to its final state. 56 •• Picture the Problem We can find the work done by the gas during this process from the area under the curve. The path from the initial state i to the final state f is shown on the PV diagram:


( )( )

kJ1.10LatmJ101.3Latm100

L05atm3atm121

trapezoidgasby

=⋅

×⋅=

+== AW

Remarks: You could use the linearity of the path connecting the initial and final states and the coordinates of the endpoints to express P as a function of V. You could then integrate this function between 1 and 3 L to find the work done by the gas as it goes from its initial to its final state. 57 •• Picture the Problem We can find the work done by the gas from the area under the PV curve provided we can find the pressure and volume coordinates of the initial and final states. We can find these coordinates by using the ideal gas law and the condition

.2APT = Apply the ideal-gas law with n = 1 mol and 2APT = to obtain:

2RAPPV = ⇒ RAPV = (1) This result tells us that the volume varies linearly with the pressure.

Solve the condition on the temperature for the pressure of the gas:

ATP 0

0 =


1389

Find the pressure when the temperature is 4T0:

000 224 P

AT

ATP ===

Using equation (1), express the coordinates of the final state:

( )00 2,2 PV

The PV diagram for the process is shown to the right:


( )( )

0023

000021

trapezoidgasby 23

VP

VVPPAW

=

−+==

*58 • Picture the Problem From the ideal gas law, PV = NkT, or V = NkT/P. Hence, on a VT diagram, isobars will be straight lines with slope 1/P. A spreadsheet program was used to plot the following graph. The graph was plotted for 1 mol of gas.

0

50

100

150

200

250

300

0 50 100 150 200 250 300 350

T (K)

V (m

3 )

P = 1 atmP = 0.5 atmP = 0.1 atm

Chapter 18

1390

59 •• Picture the Problem The PV diagram shows the isothermal expansion of the ideal gas from its initial state 1 to its final state 2. We can use the ideal-gas law for a fixed amount of gas to find V2 and then evaluate

∫ PdV for an isothermal process to find the

work done by the gas. In part (b) of the problem we can apply the first law of thermodynamics to find the heat added to the gas during the expansion.

(a) Express the work done by a gas during an isothermal process:

∫∫∫ ===2

1

2

1

2

1

11gasby

V

V

V

V

V

V VdVVP

VdVnRTdVPW

Apply the ideal-gas law for a fixed amount of gas undergoing an isothermal process:

2211 VPVP = or 1

1

1

2

PP

VV

=

Solve for and evaluate V2:

( ) L8L4kPa100kPa200

12

12 === V

PPV


( )( )

( )[ ]

( )

J555L

m10LkPa800

L4L8lnLkPa800

lnLkPa800

L4kPa200

33

L8L4

L8

L4gasby

−

×⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=

⋅=

= ∫

V

VdVW

(b) Apply the first law of thermodynamics to the system to obtain:

onintin WEQ −∆=

or, because ∆Eint = 0 for an isothermal process,

onin WQ −=

Because the work done by the gas is the negative of the work done on the gas:

( ) gasby gasby in WWQ =−−=


1391


J555in =Q

Heat Capacities of Gases and the Equipartition Theorem 60 • Picture the Problem We can find the number of moles of the gas from its heat capacity at constant volume using nRC 2

3V = . We can find the internal energy of the gas from

TCE Vint = and the heat capacity at constant pressure using nRCC += VP .

(a) Express CV in terms of the number of moles in the monatomic gas:

nRC 23

V =

Solve for n: R

Cn3

2 V=


( )( ) 99.3

KJ/mol8.3143J/K49.82

=⋅

=n

(b) Relate the internal energy of the gas to its temperature:

TCE Vint =

Substitute numerical values and evaluate Eint:

( )( ) kJ14.9K300J/K49.8int ==E

(c) Relate the heat capacity at constant pressure to the heat capacity at constant volume:

nRnRnRnRCC 25

23

VP =+=+=

Substitute numerical values and evaluate CP:

( )( )J/K82.9

KJ/mol8.3143.9925

P

=

⋅=C

61 • Picture the Problem The Dulong-Petit law gives the molar specific heat of a solid, c′. The specific heat is defined as c = c′/M where M is the molar mass. Hence we can use this definition to find M and a periodic table to identify the element. (a) Apply the Dulong-Petit law: Rc' 3= or

MRc 3

=

Chapter 18

1392

Solve for M: cRM 3

=

Substitute numerical values and evaluate M:

g/mol7.55KkJ/kg0.447KJ/mol24.9

=⋅⋅

=M

iron.likely most iselement that thesee weelements

theof tableperiodic theConsulting

*62 •• Picture the Problem The specific heats of air at constant volume and constant pressure are given by cV = CV/m and cP = CP/m and the heat capacities at constant volume and constant pressure are given by nRC 2

5V = and nRC 2

7P = , respectively.

(a) Express the specific heats per unit mass of air at constant volume and constant pressure:

mCc V

V = (1)

and

mCc P

P = (2)

Express the heat capacities of a diatomic gas in terms of the gas constant R, the number of moles n, and the number of degrees of freedom:

nRC 25

V =

and nRC 2

7P =

Express the mass of 1 mol of air:

22 ON 26.074.0 MMm +=


22 ONV 26.074.02

5MM

nRc+

=

Substitute numerical values and evaluate cV:

( )( )( ) ( )[ ] KJ/kg716

kg103226.0kg102874.02KJ/mol314.8mol15

33V ⋅=×+×⋅

= −−c


22 ONP 26.074.02

7MM

nRc+

=


1393

Substitute numerical values and evaluate cP:

( )( )( ) ( )[ ] KJ/kg1002

kg103226.0kg102874.02KJ/mol314.8mol17

33P ⋅=×+×⋅

= −−c

(b) Express the percent difference between the value from the Handbook of Chemistry and Physics and the calculated value:

%91.2KJ/g1.032

K1.002J/gKJ/g1.032difference% =⋅

⋅−⋅=

63 •• Picture the Problem We know that, during a constant-volume process, no work is done and that we can calculate the heat added during this expansion from the heat capacity at constant volume and the change in the absolute temperature. We can then use the first law of thermodynamics to find the change in the internal energy of the gas. In part (b), we can proceed similarly; using the heat capacity at constant pressure rather than constant volume. (a) The increase in the internal energy of the ideal diatomic gas is given by:

TnRE ∆=∆ 25

int

Substitute numerical values and evaluate ∆Eint:

( )( )( )kJ24.6

K300KJ/mol315.8mol125

int

=

⋅=∆E

For a constant-volume process: 0on =W

From the 1st law of thermodynamics we have:

onintin WEQ −∆=


kJ6.240kJ24.6in =−=Q

(b) Because ∆Eint depends only on the temperature difference:

kJ24.6int =∆E

Relate the heat added to the gas to its heat capacity at constant pressure and the change in its temperature:

( ) TnRTnRnRTCQ ∆=∆+=∆= 27

25

Pin

Chapter 18

1394


( )( )( )kJ73.8


in

=

⋅=Q

Apply the first law of thermodynamics to find W: kJ2.49

kJ6.24kJ73.8ininton

=

−=−∆= QEW

(c) Integrate dWon = P dV to obtain: ( ) ( )ififon

f

i

TTnRVVPPdVWV

V

−=−== ∫

Substitute numerical values and evaluate Won:

( )( )( )kJ2.49

K300KJ/mol8.314mol1on

=

⋅=W

64 •• Picture the Problem Because this is a constant-volume process, we can use

TCQ ∆= V to express Q in terms of the temperature change and the ideal-gas law for a

fixed amount of gas to find ∆T. Express the amount of heat Q that must be transferred to the gas if its pressure is to triple:

( )0f25

V

TTnRTCQ

−=

∆=

Using the ideal-gas law for a fixed amount of gas, relate the initial and final temperatures, pressures and volumes:

f

0

0

0 3T

VPTVP=

Solve for Tf:

0f 3TT =

Substitute and simplify to obtain: ( ) ( ) VPnRTTnRQ 00025 552 ===

65 •• Picture the Problem Let the subscripts i and f refer to the initial and final states of the gas, respectively. We can use the ideal-gas law for a fixed amount of gas to express V′ in terms of V and the change in temperature of the gas when 13,200 J of heat are transferred to it. We can find this change in temperature using TCQ ∆= P .

Using the ideal-gas law for a fixed amount of gas, relate the initial and f

f

i

i

TV'P

TVP=


1395

final temperatures, volumes, and pressures: Because the process is isobaric, we can solve for V′ to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛ ∆+=

∆+==

ii

i

i

f 1TTV

TTTV

TTVV'

Relate the heat transferred to the gas to the change in its temperature:

TnRTCQ ∆=∆= 27

P

Solve ∆T: nRQT

72

=∆

Substitute to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

i721nRT

QVV'

One mol of gas at STP occupies 22.4 L. Substitute numerical values and evaluate V′:

( ) ( )( )( )( ) L6.59

K273KJ/mol8.314mol17kJ13.221m1022.4 33 =⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

+×= −V'

66 •• Picture the Problem We can use the relationship between CP and CV ( nRCC += VP )

to find the number of moles of this particular gas. In parts (b) and (c) we can use the number of degrees of freedom associated with monatomic and diatomic gases, respectively, to find CP and CV. (a) Express the heat capacity of the gas at constant pressure to its heat capacity at constant volume:

nRCC += VP

Solve for n: R

CCn VP −=


mol3.50KJ/mol8.314

J/K29.1=

⋅=n

(b) CV for a monatomic gas is given by:

nRC 23

V =

Chapter 18

1396

Substitute numerical values and evaluate CV:

( )( )J/K43.6

KJ/mol8.314mol3.523

V

=

⋅=C

Express CP for a monatomic gas:

nRC 25

P =

Substitute numerical values and evaluate CP:

( )( )J/K7.27

KJ/mol8.314mol3.525

P

=

⋅=C

(c) If the diatomic molecules rotate but do not vibrate they have 5 degrees of freedom:

( )( )J/K7.27

KJ/mol8.314mol3.525

25

V

=

⋅== nRC

and ( )( )

J/K102

KJ/mol8.314mol3.527

27

P

=

⋅== nRC

*67 •• Picture the Problem We can find the change in the heat capacity at constant pressure as CO2 undergoes sublimation from the energy per molecule of CO2 in the solid and gaseous states. Express the change in the heat capacity (at constant pressure) per mole as the CO2 undergoes sublimation:

solidP,gasP,P CCC −=∆

Express Cp,gas in terms of the number of degrees of freedom per molecule:

( ) NkNkfC 25

21

gasP, == because each molecule has three translational and two rotational degrees of freedom in the gaseous state.

We know, from the Dulong-Petit Law, that the molar specific heat of most solids is 3R = 3Nk. This result is essentially a per-atom result as it was obtained for a monatomic solid with six degrees of freedom. Use this result and the fact CO2 is triatomic to express CP,solid:

NkNkC 9atoms3atom3

solidP, =×=

Substitute to obtain: NkNkNkC 213

218

25

P −=−=∆


1397

68 •• Picture the Problem We can find the initial internal energy of the gas from

nRTU 23

i = and the final internal energy from the change in internal energy resulting

from the addition of 500 J of heat. The work done during a constant-volume process is zero and the work done during the constant-pressure process can be found from the first law of thermodynamics. (a) Express the initial internal energy of the gas in terms of its temperature:

nRTE 23

iint, =

Substitute numerical values and evaluate Eint,i:

( )( )( )kJ3.40


iint,

=

⋅=E

(b) Relate the final internal energy of the gas to its initial internal energy:

TCEEEE ∆+=∆+= Viint,intiint,fint,

Express the change in temperature of the gas resulting from the addition of heat:

P

in

CQT =∆

Substitute to obtain: in

P

Viint,fint, Q

CCEE +=

Substitute numerical values and evaluate Eint,f:

( ) kJ70.3J500kJ40.32523

fint, =+=nRnRE

(c) Relate the final internal energy of the gas to its initial internal energy:

intiint,fint, EEE ∆+=

Apply the first law of thermodynamics to the constant-volume process:

oninint WQE +=∆

or, because Won = 0, J500inint ==∆ QE

Substitute numerical values and evaluate Eint,f:

kJ3.90J500kJ.403fint, =+=E

Chapter 18

1398

69 •• Picture the Problem We can use ( )NkfC 2

1waterV, = to express CV,water and then count

the number of degrees of freedom associated with a water molecule to determine f. Express CV,water in terms of the number of degrees of freedom per molecule:

( )NkfC 21

waterV, = where f is the number of degrees of freedom associated with a water molecule.

atom).per (2freedom of degrees 4 additionalan in resulting atom,oxygen eagainst th

ecan vibrat atomshydrogen theofeach addition,In freedom. of degrees rotational threeand freedom of degrees onal translati threeare There

Substitute for f to obtain: ( ) NkNkC 510 2

1waterV, ==

Quasi-Static Adiabatic Expansion of a Gas *70 •• Picture the Problem The adiabatic expansion is shown in the PV diagram. We can use the ideal-gas law to find the initial volume of the gas and the equation for a quasi-static adiabatic process to find the final volume of the gas. A second application of the ideal-gas law, this time at the final state, will yield the final temperature of the gas. In part (c) we can use the first law of thermodynamics to find the work done by the gas during this process.

(a) Apply the ideal-gas law to express the initial volume of the gas:

i

ii P

nRTV =

Substitute numerical values and evaluate Vi:

( )( )( )

L2.24m102.24atm

kPa101.3atm10

K273KJ/mol8.314mol1

33

i

=×=

×

⋅=

−

V


1399

Use the relationship between the pressures and volumes for a quasi-static adiabatic process to express Vf:

γγffii VPVP = ⇒

γ1

f

iif ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

Substitute numerical values and evaluate Vf: ( )

L5.88

atm2atm10L2.24

531

f

iif

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ

PPVV

(b) Apply the ideal-gas law to express the final temperature of the gas:

nRVPT ff

f =

Substitute numerical values and evaluate Tf:

( )( )

K143

Katm/molL108.206L5.88atm2

2f

=

⋅⋅×= −T

(c) Apply the first law of thermodynamics to express the work done on the gas:

ininton QEW −∆=

or, because the process is adiabatic, TnRTCEW ∆=∆=∆= 2

3Vinton


( )( )( )kJ1.62


on

−=

−⋅=W

Because Wby the gas = −Won: kJ1.62gasby =W

71 • Picture the Problem We can use the temperature-volume equation for a quasi-static adiabatic process to express the final temperature of the gas in terms of its initial temperature and the ratio of its heat capacitiesγ. Because nRCC += VP , we can

determine γ for each of the given heat capacities at constant volume. Express the temperature-volume relationship for a quasi-static adiabatic process:

1-ff

1-ii

γγ VTVT =

Solve for the final temperature: ( ) 1

i

1

i21

ii

1

f

iif 2 −

−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= γ

γγ

TV

VTVVTT

Chapter 18

1400

(a) Evaluate γ for nR23

VC = : 35

2325

V

P ===nRnR

CCγ

Evaluate Tf: ( )( ) K4652K293 1

f35

== −T

(b) Evaluate γ for nR25

VC = : 57

2527

V

P ===nRnR

CCγ

Evaluate Tf: ( )( ) K8732K293 1

f57

== −T

72 • Picture the Problem We can use the temperature-volume and pressure-volume equations for a quasi-static adiabatic process to express the final temperature and pressure of the gas in terms of its initial temperature and pressure and the ratio of its heat capacities. Express the temperature-volume relationship for a quasi-static adiabatic process:

1ff

1ii

−− = γγ VTVT

Solve for the final temperature: ( ) 1

i

1

i41

ii

1

f

iif 4 −

−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= γ

γγ

TV

VTVVTT

Using γ = 5/3 for neon, evaluate Tf: ( )( ) K3874K293 1

f35

== −T

Express the relationship between the pressures and volumes for a quasi-static adiabatic process:

γγffii VPVP =

Solve for Pf: γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

i41

iif V

VPP

Substitute numerical values and evaluate Pf:

( )( ) atm10.14atm1 35f ==P

*73 •• Picture the Problem We can use the ideal-gas law to find the initial volume of the gas. In part (a) we can apply the ideal-gas law for a fixed amount of gas to find the final volume and the expression (Equation 19-16) for the work done in an isothermal process. Application of the first law of thermodynamics will allow us to find the heat absorbed by the gas during this process. In part (b) we can use the relationship between the pressures


1401

and volumes for a quasi-static adiabatic process to find the final volume of the gas. We can apply the ideal-gas law to find the final temperature and, as in (a), apply the first law of thermodynamics, this time to find the work done by the gas. Use the ideal-gas law to express the initial volume of the gas:

i

ii P

nRTV =


( )( )( )

L3.12m103.12kPa400

K300KJ/mol8.314mol0.5

33

i

=×=

⋅=

−

V

(a) Because the process is isothermal:

K300if == TT

Use the ideal-gas law for a fixed amount of gas to express Vf: f

ff

i

ii

TVP

TVP

=

or, because T = constant,

f

iif P

PVV =

Substitute numerical values and evaluate Vf:

( ) L7.80kPa160kPa400L3.12f =⎟⎟

⎠

⎞⎜⎜⎝

⎛=V

Express the work done by the gas during the isothermal expansion:

i

fgasby ln

VVnRTW =


( )( )

( )

kJ14.1

L3.12L7.80lnK300

KJ/mol8.314mol0.5gasby

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

⋅=W

Noting that the work done by the gas during the process equals the negative of the work done on the gas, apply the first law of thermodynamics to find the heat absorbed by the gas:

( )kJ1.14

kJ1.140onintin

=

−−=−∆= WEQ

Chapter 18

1402

(b) Using γ = 5/3 and the relationship between the pressures and volumes for a quasi-static adiabatic process, express Vf:

γγffii VPVP = ⇒

γ1

f

iif ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

Substitute numerical values and evaluate Vf: ( ) L5.41

kPa160kPa400L12.3

53

f =⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Apply the ideal-gas law to find the final temperature of the gas:

nRVPT ff

f =


( )( )( )( )

K208

KJ/mol8.314mol0.5m105.41kPa160 33

f

=

⋅×

=−

T

For an adiabatic process: 0in =Q

Apply the first law of thermodynamics to express the work done on the gas during the adiabatic process:

TnRTCQEW ∆=−∆=−∆= 23

Vininton 0


( )( )( )

J745K300K208

KJ/mol8.314mol0.523

on

−=−×

⋅=W

Because the work done by the gas equals the negative of the work done on the gas:

( ) J745J574gasby =−−=W

74 •• Picture the Problem We can use the ideal-gas law to find the initial volume of the gas. In part (a) we can apply the ideal-gas law for a fixed amount of gas to find the final volume and the expression (Equation 19-16) for the work done in an isothermal process. Application of the first law of thermodynamics will allow us to find the heat absorbed by the gas during this process. In part (b) we can use the relationship between the pressures and volumes for a quasi-static adiabatic process to find the final volume of the gas. We can apply the ideal-gas law to find the final temperature and, as in (a), apply the first law of thermodynamics, this time to find the work done by the gas.


1403

Use the ideal-gas law to express the initial volume of the gas:

i

ii P

nRTV =


( )( )( )

L3.12m103.12kPa400

K300KJ/mol8.314mol0.5

33

i

=×=

⋅=

−

V

(a) Because the process is isothermal:

K300if == TT

Use the ideal-gas law for a fixed amount of gas to express Vf: f

ff

i

ii

TVP

TVP

=

or, because T = constant,

f

iif P

PVV =


( ) L7.80kPa160kPa400L3.12f =⎟⎟

⎠

⎞⎜⎜⎝

⎛=V

Express the work done by the gas during the isothermal expansion:

i

fgasby ln

VVnRTW =


( )( )

( )

kJ14.1

L3.12L7.80lnK300

KJ/mol8.314mol0.5gasby

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

⋅=W

Noting that the work done by the gas during the isothermal expansion equals the negative of the work done on the gas, apply the first law of thermodynamics to find the heat absorbed by the gas:

( )kJ1.14

kJ1.140onintin

=

−−=−∆= WEQ

(b) Using γ = 1.4 and the relationship between the pressures and volumes for a quasi-static adiabatic process, express Vf:

γγffii VPVP = ⇒

γ1

f

iif ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

Chapter 18

1404

Substitute numerical values and evaluate Vf: ( ) L00.6

kPa160kPa400L12.3

1.41

f =⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Apply the ideal-gas law to express the final temperature of the gas:

nRVPT ff

f =


( )( )( )( )

K231

KJ/mol8.314mol0.5m106kPa160 33

f

=

⋅×

=−

T

For an adiabatic process: 0in =Q

Apply the first law of thermodynamics to express the work done on the gas during the adiabatic expansion:

TnRTCQEW ∆=−∆=−∆= 25

Vininton 0


( )( )( )

J717K300K312

KJ/mol8.314mol0.525

on

−=−×

⋅=W

Noting that the work done by the gas during the adiabatic expansion is the negative of the work done on the gas, we have:

( ) J717J717gasby =−−=W

75 •• Picture the Problem We can eliminate the volumes from the equations relating the temperatures and volumes and the pressures and volumes for a quasi-static adiabatic process to obtain a relationship between the temperatures and pressures. We can find the initial volume of the gas using the ideal-gas law and the final volume using the pressure-volume relationship. In parts (d) and (c) we can find the change in the internal energy of the gas from the change in its temperature and use the first law of thermodynamics to find the work done by the gas during its expansion. (a) Express the relationship between temperatures and volumes for a quasi-static adiabatic process:

1ff

1ii

−− = γγ VTVT


1405

Express the relationship between pressures and volumes for a quasi-static adiabatic process:

γγffii VPVP = (1)

Eliminate the volume between these two equations to obtain:

γ11

i

fif

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

PPTT

Substitute numerical values and evaluate Tf: ( ) K263

atm5atm1K500

3511

f =⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

T

(b) Solve equation (1) for Vf: γ

1

f

iif ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

Apply the ideal-gas law to express Vi:

i

ii P

nRTV =


( )( )( )

L4.10atm

kPa101.35atm

K500KJ/mol8.314mol0.5i

=

×

⋅=V

Substitute for Vi and evaluate Vf:

( ) L8.10atm1atm5L10.4

53

f =⎟⎟⎠

⎞⎜⎜⎝

⎛=V

(d) Relate the change in the internal energy of the helium gas to the change in its temperature:

TnRTCE ∆=∆=∆ 23

Vint


( )( )( )

kJ48.1

K500K263KJ/mol8.314mol0.52

3int

−=

−×⋅=∆E

(c) Use the first law of thermodynamics to express the work done on the gas:

intintininton 0 EEQEW ∆=−∆=−∆=


kJ48.1on −=W

Chapter 18

1406

Because the work done by the gas equals the negative of Won:

( )kJ48.1

kJ48.1ongasby

=

−−=−= WW

*76 ••• Picture the Problem Consider the process to be accomplished in a single compression. The initial pressure is 1 atm = 101 kPa. The final pressure is (101 + 482) kPa = 583 kPa, and the final volume is 1 L. Because air is a mixture of diatomic gases, γair = 1.4. We can find the initial volume of the air using γγ

ffii VPVP = and use Equation 19-39 to find the

work done by the air. Express the work done in an adiabatic process:

1ffii

−−

=γ

VPVPW (1)

Use the relationship between pressure and volume for a quasi-static adiabatic process to express the initial volume of the air:

γγffii VPVP = ⇒

γ1

i

ffi ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

Substitute numerical values and evaluate Vi: ( ) L50.3

kPa101kPa583L1

4.11

i =⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Substitute numerical values in equation (1) and evaluate W:

( )( ) ( )( ) J57414.1

m10kPa835m103.5kPa101 3333

−=−−×

=−−

W

where the minus sign tells us that work is done on the gas. 77 ••• Picture the Problem We can integrate PdV using the equation of state for an adiabatic process to obtain Equation 18-39. Express the work done by the gas during this adiabatic expansion:

∫=2

1

gasby

V

V

PdVW

For an adiabatic process: CPV == constantγ (1) and

γ−= CVP

Substitute and evaluate the integral to obtain: ( )γγγ

γ−−− −

−== ∫ 1

112gasby 1

2

1

VVCdVVCWV

V


1407

From equation (1) we have: γγ22

12 VPCV =− and γγ

111

1 VPCV =−


1122111122

gasby −−

=−−

=γγ

γγγγ VPVPVPVPW ,

which is Equation 18-39. Cyclic Processes 78 •• Picture the Problem To construct the PV diagram we’ll need to determine the volume occupied by the gas at the beginning and ending points for each process. Let these points be A, B, C, and D. We can apply the ideal-gas law to the starting point (A) to find VA. To find the volume at point B, we can use the relationship between pressure and volume for a quasi-static adiabatic process. We can use the ideal-gas law to find the volume at point C and, because they are equal, the volume at point D. We can apply the first law of thermodynamics to find the amount of heat added to or subtracted from the gas during the complete cycle. (a) Using the ideal-gas law, express the volume of the gas at the starting point A of the cycle:

A

AA P

nRTV =

Substitute numerical values and evaluate VA:

( )( )( )

L4.81atm

kPa101.3atm5

K293KJ/mol8.314mol1A

=

×

⋅=V

Use the relationship between pressure and volume for a quasi-static adiabatic process to express the volume of the gas at point B; the end point of the adiabatic expansion:

γ1

B

AAB ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

PPVV

Substitute numerical values and evaluate VB: ( ) L2.15

atm1atm5L81.4

4.11

B =⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Using the ideal-gas law for a fixed amount of gas, express the volume occupied by the gas at points C and D:

C

CDC P

nRTVV ==

Chapter 18

1408

Substitute numerical values and evaluate VC:

( )( )( )

L0.24atm

kPa101.3atm1

K293KJ/mol8.314mol1C

=

×

⋅=V

The complete cycle is shown in the diagram.

(b) Note that for the paths A→B and B→C, Wby gas, the work done by the gas, is positive. For the path D→A, Wby gas is negative, and greater in magnitude than WA→C. Therefore the total work done by the gas is negative. Find the area enclosed by the cycle by noting that each rectangle of dotted lines equals 5 atm⋅L and counting the rectangles:

( )( ) ( )

kJ6.58

LatmJ101.3Latm65eL/rectanglatm5rectangles13gasby

−=

⎟⎠⎞

⎜⎝⎛

⋅⋅−=⋅−≈W

(c) The work done on the gas equals the negative of the work done by the gas. Apply the first law of thermodynamics to find the amount of heat added to or subtracted from the gas during the complete cycle:

( )kJ6.58

kJ6.580onintin

=

−−=−∆= WEQ

because ∆Eint = 0 for the complete cycle.

(d) Express the work done during the complete cycle:

ADDCCBBA →→→→ +++= WWWWW

Because A→B is an adiabatic process: 1

BBAABA −

−=→ γ

γγ VPVPW


1409

Substitute numerical values and evaluate WA→B:

( )( ) ( )( )

( )

kJ25.2LatmJ101.3Latm3.22

14.1L2.51atm1L4.82atm5

BA

=

⎟⎠⎞

⎜⎝⎛

⋅⋅=

−−

=→W

B→C is an isobaric process:

( )( )

( )


L15.2L24.0atm1CB

=

⎟⎠⎞

⎜⎝⎛

⋅⋅=

−=∆=→ VPW

C→D is a constant-volume process:

0DC =→W

D→A is an isobaric process: ( )( )

( )


L42L5atm5AD

−=

⎟⎠⎞

⎜⎝⎛

⋅⋅−=

−=∆=→ VPW


kJ6.48

kJ9.620kJ0.8912.25kJ

−=

−++=W

Note that our result in part (b) agrees with this more accurate value to within 2%.

*79 •• Picture the Problem The total work done as the gas is taken through this cycle is the area bounded by the two processes. Because the process from 1→2 is linear, we can use the formula for the area of a trapezoid to find the work done during this expansion. We can use ( )ifprocess isothermal ln VVnRTW = to find the work done on the gas during the

process 2→1. The total work is then the sum of these two terms. Express the net work done per cycle:

1221total →→ += WWW (1)

Work is done by the gas during its expansion from 1 to 2 and hence is equal to the negative of the area of the trapezoid defined by this path and the vertical lines at V1 = 11.5 L and V2 = 23 L. Use the formula for the area of a trapezoid to express

( )( )atmL3.17

atm1atm2L11.5L2321

trap21

⋅−=+−−=

−=→ AW

Chapter 18

1410

W1→2: Work is done on the gas during the isothermal compression from V2 to V1 and hence is equal to the area under the curve representing this process. Use the expression for the work done during an isothermal process to express W2→1:

⎟⎟⎠

⎞⎜⎜⎝

⎛=→

i

f12 ln

VVnRTW

Apply the ideal-gas law at point 1 to find the temperature along the isotherm 2→1:

( )( )( )( ) K280

Katm/molL10206.8mol1L5.11atm2

2 =⋅⋅×

== −nRPVT

Substitute numerical values and evaluate W2→1:

( )( )( ) atmL9.15L23L5.11lnK280Katm/molL10206.8mol1 2

12 ⋅=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅×= −

→W

Substitute in equation (1) and evaluate Wnet:

J142atmL

J101.325atmL40.1

atmL15.9atmL3.17net

−=⋅

×⋅−=

⋅+⋅−=W

Remarks: The work done by the gas during each cycle is 142 J. 80 •• Picture the Problem We can apply the ideal-gas law to find the temperatures T1, T2, and T3. We can use the appropriate work and heat equations to calculate the heat added and the work done by the gas for the isothermal process (1→2), the constant-volume process (2→3), and the isobaric process (3→1).


1411

(a) The cycle is shown in the diagram:

(c) Use the ideal-gas law to find T1:

( )( )( )( )

K4.42

Katm/molL108.206mol2L2atm2

2

111

=

⋅⋅×=

=

−

nRVPT

Because the process 1→2 is isothermal:

K4.422 =T

Use the ideal-gas law to find T3:

( )( )( )( )

K7.84


2

333

=

⋅⋅×=

=

−

nRVPT

(b) Because the process 1→2 is isothermal, Qin,1→2 = Wby gas,1→2:

⎟⎟⎠

⎞⎜⎜⎝

⎛== →→

1

221gas,by 21 in, ln

VVnRTWQ

Substitute numerical values and evaluate Qin,1→2:

( )( )( ) J281L2L4lnK24.4KJ/mol8.314mol221 in, =⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=→Q

Because process 2→3 takes place at constant volume:

032 =→W

Because process 2→3 takes place at constant volume, Won,2→3 = 0, and:

( )2323

V3int,23in,2 TTnRTCEQ −=∆=∆= →→

Substitute numerical values and evaluate Qin,2→3:

Chapter 18

1412

( )( )( ) J606K4.42K48.7KJ/mol8.314mol223

3in,2 =−⋅=→Q

Because process 3→1 is isobaric: )( 312

5P13 TTnRTCQ −=∆=→

Substitute numerical values and evaluate Q3→1:

( )( )( ) kJ01.1K7.84K4.24KJ/mol8.314mol225

13 −=−⋅=→Q

The work done by the gas from 3 to 1 equals the negative of the work done on the gas:

( )313,13,11gas,3by VVPVPW −=∆−=→

Substitute numerical values and evaluate Wby gas,3→2:

( )( )

( )

J405

LatmJ101.3Latm4

L4L2atm21gas,3by

=

⎟⎠⎞

⎜⎝⎛

⋅⋅−−=

−−=→W

81 ••• Picture the Problem We can find the temperatures, pressures, and volumes at all points for this ideal monatomic gas (3 degrees of freedom) using the ideal-gas law and the work for each process by finding the areas under each curve. We can find the heat exchanged for each process from the heat capacities and the initial and final temperatures for each process. Express the total work done by the gas per cycle:

DCCBBAADtotgas,by →→→→ +++= WWWWW

1. Use the ideal-gas law to find the volume of the gas at point D: ( )( )( )

( )( )L29.5

kPa/atm101.3atm2K360KJ/mol8.314mol2

D

DD

=

⋅=

=P

nRTV

2. We’re given that the volume of the gas at point B is three times that at point D:

L6.883 DCB

=== VVV

Use the ideal-gas law to find the pressure of the gas at point C:


1413

( )( )( ) atm667.0L6.88

K360Katm/molL10206.8mol2 2

C

CC =

⋅⋅×==

−

VnRTP

We’re given that the pressure at point B is twice that at point C:

( ) atm33.1atm667.022 CB === PP

3. Because path DC represents an isothermal process:

K360CD == TT

Use the ideal-gas law to find the temperatures at points B and A: ( )( )

( )( )K720

Katm/molL108.206mol2L88.6atm1.333

2

BBBA

=⋅⋅×

=

==

−

nRVPTT

Because the temperature at point A is twice that at D and the volumes are the same, we can conclude that:

atm42 DA == PP

The pressure, volume, and temperature at points A, B, C, and D are summarized in the table to the right.

Point P V T

(atm) (L) (K) A 4 29.5 720 B 1.33 88.6 720 C 0.667 88.6 360 D 2 29.5 360

4. For the path D→A, 0AD =→W

and: ( )DA23

AD23

AD int,AD

TTnRTnREQ

−=

∆=∆= →→→

Substitute numerical values and evaluate QD→A:

( )( )( ) kJ98.8K360K720KJ/mol8.314mol223

AD =−⋅=→Q

For the path A→B:

⎟⎟⎠

⎞⎜⎜⎝

⎛== →→

A

BBA,BABA ln

VVnRTQW

Substitute numerical values and evaluate WA→B:

( )( )( ) kJ2.13L29.5L88.6lnK720KJ/mol8.314mol2BA =⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=→W

Chapter 18

1414

and, because process A→B is isothermal, 0BA int, =∆ →E

For the path B→C, 0CB =→W , and: ( )BC2

3VCBCB TTnRTCUQ −=∆=∆= →→

Substitute numerical values and evaluate QB→C:

( )( )( ) kJ98.8K720K360KJ/mol8.314mol223

CB −=−⋅=→Q

For the path C→D:

⎟⎟⎠

⎞⎜⎜⎝

⎛=→

C

DDC,DC ln

VVnRTW

Substitute numerical values and evaluate WC→D:

( )( )( ) kJ58.6L6.88L5.92lnK603KJ/mol8.314mol2DC −=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=→W

Also, because process A→B is isothermal, 0BAint, =∆ →E , and

kJ58.6DCDC −== →→ WQ

Qin, Won, and ∆Eint are summarized for each of the processes in the table to the right.

Process Qin Won ∆Eint

(kJ) (kJ) (kJ) D→A 98.8 0 8.98

A→B 2.13 −13.2

0

B→C 98.8− 0 −8.98

C→D 58.6− 6.58 0

Referring to the table, find the total work done by the gas per cycle:

kJ6.62

kJ6.580kJ13.20DCCBBAADtot

=

−++=+++= →→→→ WWWWW

Remarks: Note that, as it should be, ∆Eint is zero for the complete cycle. *82 ••• Picture the Problem We can find the temperatures, pressures, and volumes at all points for this ideal diatomic gas (5 degrees of freedom) using the ideal-gas law and the work for each process by finding the areas under each curve. We can find the heat exchanged for


1415

each process from the heat capacities and the initial and final temperatures for each process. Express the total work done by the gas per cycle:

DCCBBAADtotgas,by →→→→ +++= WWWWW

1. Use the ideal-gas law to find the volume of the gas at point D: ( )( )( )

( )( )L29.5

kPa/atm101.3atm2K360KJ/mol8.314mol2

D

DD

=

⋅=

=P

nRTV

2. We’re given that the volume of the gas at point B is three times that at point D:

L6.883 DCB

=== VVV

Use the ideal-gas law to find the pressure of the gas at point C:

( )( )( ) atm667.0L6.88


C

CC =

⋅⋅×==

−

VnRTP

We’re given that the pressure at point B is twice that at point C:

( ) atm33.1atm667.022 CB === PP

3. Because path DC represents an isothermal process:

K360CD == TT

Use the ideal-gas law to find the temperatures at points B and A: ( )( )

( )( )K720

Katm/molL108.206mol2L88.6atm1.333

2

BBBA

=⋅⋅×

=

==

−

nRVPTT

Because the temperature at point A is twice that at D and the volumes are the same, we can conclude that:

atm42 DA == PP

The pressure, volume, and temperature at points A, B, C, and D are summarized in the table to the right.

Point P V T (atm) (L) (K)

A 4 29.5 720

Chapter 18

1416

B 1.33 88.6 720 C 0.667 88.6 360 D 2 29.5 360

4. For the path D→A, 0AD =→W and:

( )

( )( )( )kJ0.15

K360K720KJ/mol8.314mol225

DA25

AD25

ADAD

=

−⋅=

−=∆=∆= →→→ TTnRTnRUQ

For the path A→B:

( )( )( )

kJ2.13L29.5L88.6lnK720KJ/mol8.314mol2ln

A

BBA,BABA

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛== →→ V

VnRTQW

and, because process A→B is isothermal, 0BAint, =∆ →E

For the path B→C, 0CB =→W and:

( ) ( )( )( )

kJ0.15K720K360KJ/mol8.314mol22

5BC2

5VCBCB

−=

−⋅=−=∆=∆= →→ TTnRTCUQ

For the path C→D:

( )( )( ) kJ58.6L6.88L5.92lnK603KJ/mol8.314mol2ln

C

DDC,DC −=⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=→ V

VnRTW

Also, because process A→B is isothermal, 0BAint, =∆ →E and

kJ58.6DCDC −== →→ WQ

Qin, Won, and ∆Eint are summarized for each of the processes in the table to the right.

Process Qin Won ∆Eint

(kJ) (kJ) (kJ) D→A 0.15 0 15.0

A→B 2.13 −13.2 0

B→C 0.15− 0 −15.0

C→D 58.6− 6.58 0


1417

Referring to the table and noting that the work done by the gas equals the negative of the work done on the gas, find the total work done by the gas per cycle:

kJ6.62kJ6.580kJ13.20DCCBBAADtotgas,by =−++=+++= →→→→ WWWWW

Remarks: Note that ∆Eint for the complete cycle is zero and that the total work done is the same for the diatomic gas of this problem and the monatomic gas of problem 81. 83 ••• Picture the Problem We can use the equations of state for adiabatic and isothermal processes to express the work done on or by the system, the heat entering or leaving the system, and the change in internal energy for each of the four processes making up the Carnot cycle. We can use the first law of thermodynamics and the definition of the efficiency of a Carnot cycle to show that the efficiency is 1 – Qc / Qh. (a) The cycle is shown on the PV diagram to the right:

(b) Because the process 1→2 is isothermal:

021int, =∆ →E

Apply the first law of thermodynamics to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛=== →→

1

2h2121h ln

VVnRTWQQ

(c) Because the process 3→4 is isothermal:

043 =∆ →U

Chapter 18

1418


⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛=== →→

4

3c

3

4c4343c

ln

ln

VVnRT

VVnRTWQQ

where the minus sign tells us that heat is given off by the gas during this process.

(d) Apply the equation for a quasi-static adiabatic process at points 4 and 1 to obtain:

11h

14c

−− = γγ VTVT

Solve for the ratio V1/V4:

11

h

c

4

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γ

TT

VV

(1)

Apply the equation for a quasi-static adiabatic process at points 2 and 3 to obtain:

13c

12h

−− = γγ VTVT

Solve for the ratio V2/V3:

11

h

c

3

2−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γ

TT

VV

(2)

Equate equations (1) and (2) and rearrange to obtain:

1

2

4

3

VV

VV

=

(e) Express the efficiency of the Carnot cycle:

hQW

=ε


( ) ( )chch

incycle int,on

0 QQQQ

QEW

−−=−−=

−∆=

because Eint is a state function and 0cycle int, =∆E .


h

c

h

ch

h

on

h

gas by the

1QQ

QQQ

QW

QW

−=−

=

−==ε


1419

(f) In part (b) we established that: ⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

2hh ln

VVnRTQ

In part (c) we established that the heat leaving the system along the path 3→4 is given by:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

4

3cc ln

VVnRTQ


h

c

1

2h

4

3c

h

c

ln

ln

TT

VVnRT

VVnRT

QQ

=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

because1

2

4

3

VV

VV

= .

Remarks: This last result establishes that the efficiency of a Carnot cycle is also

given byh

cC T

Tε −= 1 .

General Problems 84 • Picture the Problem The isobaric process is shown on the PV diagram. We can express the heat that must be supplied to gas in terms of its heat capacity at constant pressure and the change in its temperature and then use the ideal-gas law for a fixed amount of gas to relate the final temperature to the initial temperature.

Relate Qin to CP and ∆T:

( ) ( )if25

ifPPin TTnRTTCTCQ −=−=∆=

Use the ideal-gas law for a fixed amount of gas to relate the initial and final volumes, pressures, and temperatures:

f

ff

i

ii

TVP

TVP

=

or, because the process is isobaric,

f

f

i

i

TV

TV

=

Solve for Tf:

iiii

ff 4

L50L200 TTT

VVT ===

Chapter 18

1420

Substitute to obtain: ( ) i215

ii25

in 34 nRTTTnRQ =−=


( )( )( )kJ56.1


in

=

⋅=Q

85 • Picture the Problem We can use the first law of thermodynamics to relate the heat removed from the gas to the work done on the gas. Apply the first law of thermodynamics to this process:

ononintin WWEQ −=−∆=

because ∆Eint = 0 for an isothermal process.

Substitute numerical values to obtain: kJ180in −=Q

Because Qremoved = −Qin: kJ180removed =Q

*86 • Picture the Problem We can find the number of moles of the gas from the expression for the work done on or by a gas during an isothermal process. Express the work done on the gas during the isothermal process:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

i

flnVVnRTW

Solve for n:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

i

flnVVRT

Wn

Substitute numerical values and evaluate n: ( )( )

mol45.9

51lnK293KJ/mol8.314

kJ180

=

⎟⎠⎞

⎜⎝⎛⋅

−=n

87 • Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process EDC is isobaric, we can find the area under this line geometrically and the first law of thermodynamics to find QAEC.


1421

(a) Using the ideal-gas law, find the temperature at point A:

( )( )( )( )

K65.2


2

AAA

=

⋅⋅×=

=

−

nRVPT

Using the ideal-gas law, find the temperature at point C: ( )( )

( )( )K81.2


2

CCC

=

⋅⋅×=

=

−

nRVPT

(b) Express the work done by the gas along the path AEC:

( )( )

kJ1.62atmL

J101.3atmL16.0

L4.01L20atm10 ECECECAEAEC

=⋅

×⋅=

−=∆+=+= VPWWW

(c) Apply the first law of thermodynamics to express QAEC: ( )ATTnRW

TnRWTCWEWQ

−+=∆+=

∆+=∆+=

C23

AEC

23

AEC

VAECintAECAEC

Substitute numerical values and evaluate QAEC:

( )( )( ) kJ2.22K65.2K81.2KJ/mol8.314mol3kJ1.62 23

AEC =−⋅+=Q

Remarks The difference between WAEC and QAEC is the change in the internal energy ∆Eint,AEC during this process. 88 •• Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process AB is isobaric, we can find the area under this line geometrically. We can use the expression for the work done during an isothermal expansion to find the work done between B and C and the first law of thermodynamics to find QABC.

Chapter 18

1422

(a) Using the ideal-gas law, find the temperature at point A:

( )( )( )( )

K65.2


2

AAA

=

⋅⋅×=

=

−

nRVPT

Use the ideal-gas law to find the temperature at point C: ( )( )

( )( )K81.2


2

CCC

=

⋅⋅×=

=

−

nRVPT

(b) Express the work done by the gas along the path ABC:

B

CBABAB

BCABABC

lnVVnRTVP

WWW

+∆=

+=

Use the ideal-gas law to find the volume of the gas at point B:

( )( )( ) L5.00atm4

K81.2Katm/molL108.206mol3 2

B

BB =

⋅⋅×==

−

PnRTV


( )( ) ( )( )( )

kJ21.3atm

J101.3atmL1.73

L5L20lnK81.2Katm/molL108.206mol3L4.01L5atm4 2

ABC

=×⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅×+−= −W

(c) Apply the first law of thermodynamics to obtain: ( )ATTnRW

TnRWTCWEWQ

−+=∆+=

∆+=∆+=

C23

AEC

23

AEC

VABCintABCABC

Substitute numerical values and evaluate QABC:

( )( )( ) kJ81.3K65.2K81.2KJ/mol8.314mol3kJ21.3 23

ABC =−⋅+=Q

Remarks: The difference between WABC and QABC is the change in the internal energy ∆Eint,ABC during this process.


1423

*89 •• Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process DC is isobaric, we can find the area under this line geometrically. We can use the expression for the work done during an isothermal expansion to find the work done between A and D and the first law of thermodynamics to find QADC. (a) Using the ideal-gas law, find the temperature at point A:

( )( )( )( )

K65.2


2

AAA

=

⋅⋅×=

=

−

nRVPT

Use the ideal-gas law to find the temperature at point C: ( )( )

( )( )K81.2


2

CCC

=

⋅⋅×=

=

−

nRVPT

(b) Express the work done by the gas along the path ADC: DCDC

A

DA

DCADADC

ln VPVV

nRT

WWW

∆+⎟⎟⎠

⎞⎜⎜⎝

⎛=

+=

Use the ideal-gas law to find the volume of the gas at point D:

( )( )( ) L1.61atm1

K65.2Katm/molL108.206mol3 2

D

DD =

⋅⋅×==

−

PnRTV

Substitute numerical values and evaluate WADC:

( )( )( )

( )( )

kJ65.2atmL

J101.325atmL2.26

L1.61L02atm1L01.4L1.61lnK65.2Katm/molL108.206mol3 2

ADC

=⋅

×⋅=

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅×= −W

(c) Apply the first law of thermodynamics to obtain: ( )ATTnRW

TnRWTCWEWQ

−+=∆+=

∆+=∆+=

C23

ADC

23

ADC

VADCintADCADC

Chapter 18

1424

Substitute numerical values and evaluate QADC:

( )( )( ) kJ25.3K65.2K81.2KJ/mol8.314mol3kJ65.2 23

ADC =−⋅+=Q

90 •• Picture the Problem We can use the ideal-gas law to find the temperatures TA and TC. Because the process AB is isobaric, we can find the area under this line geometrically. We can find the work done during the adiabatic expansion between B and C using

BCVBC TCW ∆−= and the first law of thermodynamics to find QABC.

The work done by the gas along path ABC is given by:

BC23

ABAB

BCVABAB

BCABABC

TnRVPTCVP

WWW

∆−∆=

∆−∆=+=

because, with Qin = 0, WBC = −∆Eint,BC.

Use the ideal-gas law to find TA:

( )( )( )( )

K65.2Katm/molL108.206mol3

L4.01atm42

AAA

=⋅⋅×

=

=

−

nRVPT

Use the ideal-gas law to find TB:

( )( )( )( )


L8.71atm42

BBB

=⋅⋅×

=

=

−

nRVPT

Use the ideal-gas law to find TC:

( )( )( )( )

K81.2Katm/molL108.206mol3

L02atm12

CCC

=⋅⋅×

=

=

−

nRVPT

Apply the pressure-volume relationship for a quasi-static adiabatic process to the gas at points B and C to find the volume of the gas at point B:

γγCCBB VPVP =

and


1425

( )

L71.8

L20atm4atm1 5

31

CB

CB

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= V

PPV

γ

Substitute numerical values and evaluate WABC:

( )( ) ( )( )( )

kJ18.4atm

J101.325atmL3.41

K142K81.2Katm/molL108.206mol3L4.01L71.8atm4 2

23

ABC

=×⋅=

−×⋅⋅×−−= −W

Apply the 1st law of thermodynamics to obtain: ( )ATTnRW

TnRWTCWEWQ

−+=∆+=

∆+=∆+=

C23

ABC

23

ABC

VABCintABCABC

Substitute numerical values and evaluate QABC:

( )( )( ) kJ78.4K65.2K81.2KJ/mol8.314mol3kJ18.4 23

ABC =−⋅+=Q

91 •• Picture the Problem We can find c at T = 4 K by direct substitution. Because c is a function of T, we’ll integrate dQ over the given temperature interval in order to find the heat required to heat copper from 1 to 3 K. (a) Substitute for a and b to obtain: ( )

( ) 344

2

KJ/kg1062.7KJ/kg0.0108

TTc⋅×+

⋅=−

Evaluate c at T = 4 K: ( ) ( )( )

( )( )KJ/kg1020.9

K4KJ/kg1062.7

K4KJ/kg0.0108K4

2

344

2

⋅×=

⋅×+

⋅=

−

−

c

(b) Express and evaluate the integral of Q:

( ) ( ) ( )

( ) ( ) J/kg0584.04

KJ/kg1062.72

KJ/kg0108.0

KJ/kg1062.7KJ/kg0108.0

K3

K1

444

K3

K1

22

K3

K1

344K3

K1

2f

i

=⎥⎦

⎤⎢⎣

⎡⋅×+⎥

⎦

⎤⎢⎣

⎡⋅=

⋅×+⋅==

−

− ∫∫∫

TT

dTTTdTdTTcQT

T

Chapter 18

1426

92 •• Picture the Problem We can use the first law of thermodynamics to relate the heat escaping from the system to the amount of work done by the gas and the change in its internal energy. We can use the expression for the work done during an isothermal process to find the temperature along the isotherm. Apply the first law of thermodynamics to this isothermal process:

onintin WEQ −∆=

For an isothermal process: 0int =∆E


J711cal

J4.184cal170inon

=

⎟⎠⎞

⎜⎝⎛ ×−−=−= QW

Because Wby gas = −Won:

J711gasby −=W

Express the work done during an isothermal process:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

2gasby ln

VVnRTW

Solve for T = Ti = Tf:

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

2

gasby

lnVVnR

WT

Substitute numerical values and evaluate T: ( )( )

K52.7

L18L8lnKJ/mol8.314mol2

J711

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

−=T

93 •• Picture the Problem Let the subscripts 1 and 2 refer to the initial and final values of temperature, pressure, and volume. We can relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial and final points on the path

using 1

2211

−−

=γ

VPVPW and find P1 by eliminating P2 using ,2211γγ VPVP = where, for a

diatomic gas, γ = 1.4. Once we’ve determined P1 we can use the ideal-gas law to find T1 and the first law of thermodynamics to find T2. Finally, we can apply the ideal-gas law a second time to determine P2.


1427

Relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial and final points on the path: 1

1

21

211

2211on

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

−−

=

γ

γ

VPPVP

VPVPW

Using the equation for a quasi-static adiabatic process, relate the initial and final pressures and volumes:

γγ2211 VPVP = ⇒

γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1

1

2

VV

PP


1

22

111

on −

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

=γ

γ

VVVVP

W

Solve for P1: ( )

22

11

11

VVVV

WP γγ

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=


( )( )

( )kPa6.47

L8L8L18L18

11.4J8201.41 =

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−=P

Use the ideal-gas law to find T1: ( )( )

( )( )K5.51

KJ/mol8.314mol2L18kPa47.611

1

=

⋅==

nRVPT


oninint WQE +=∆

or, because Qin = 0 for an adiabatic process, ( )122

5Vonint TTnRTCWE −=∆==∆

Solve for and evaluate T2:

( )( )K2.71

KJ/mol8.314mol2J820K.551

25

25

on12

=

⋅−

−=

−=nR

WTT

Chapter 18

1428

Use the ideal-gas law to find P2:

( )( )( )

kPa148

L8K71.2KJ/mol1438.mol2

2

22

=

⋅=

=V

nRTP

94 •• Picture the Problem Let the subscripts 1 and 2 refer to the initial and final state respectively. Because the gas is initially at STP, we know that V1 = 22.4 L, P1 = 1 atm, and T1 = 273 K. We can use ( )12ln VVnRTW −= to find the work done on the gas

during an isothermal compression. We can relate the work done on a gas during an adiabatic process to the pressures and volumes of the initial and final points on the path

using 1

2211

−−

=γ


diatomic gas, γ = 1.4. (a) Express the work done on the gas in compressing it isothermally:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

2on ln

VVnRTW

Find the number of moles in 30 g of CO (M = 28 g/mol):

mol1.07g/mol28

g30==n


( )( )( ) kJ3.9151lnK273KJ/mol8.314mol1.07on =⎟⎠⎞

⎜⎝⎛⋅−=W

(b) Express the work done on the gas in compressing it adiabatically:

1

1

21

211

2211on

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−−

−=

γ

γ

VPPVP

VPVPW


γγ2211 VPVP = ⇒

γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1

1

2

VV

PP


1429

Substitute for P2/P1and simplify to obtain:

1

2.01

1

5

12

111

1

2

1112

2

111

on −

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=γγγ

γγγ

VVVPV

VVVPV

VVVP

W


( )( )( ) ( )( ) kJ49.514.1

50.21L/mol22.4mol1.07kPa101.3 1.4

=−

−−=W

95 •• Picture the Problem Let the subscripts 1 and 2 refer to the initial and final state respectively. Because the gas is initially at STP, we know that V1 = 22.4 L, P1 = 1 atm, and T1 = 273 K. We can use ( )12ln VVnRTW −= to find the work done on the gas


using 1

2211

−−

=γ

VPVPW and find P1 by eliminating P2 using γγ2211 VPVP = . We can find γ

using the data in Table 19-3. (a) Express the work done on the gas in compressing it isothermally:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

2on ln

VVnRTW

Find the number of moles in 30 g of CO2 (M = 44 g/mol):

mol682.0g/mol44

g30==n



⎜⎝⎛⋅−=W


1

1

21

211

2211on

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−−

−=

γ

γ

VPPVP

VPVPW

Chapter 18

1430


γγ2211 VPVP = ⇒

γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1

1

2

VV

PP

Substitute for P2/P1 and simplify to obtain:

1

2.01

1

5

12

111

1

2

1112

2

111

on −

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=γγγ

γγγ

VVVPV

VVVPV

VVVP

W

From Table 18-3 we have:

Rc 39.3V =

and ( ) RRc 41.402.139.3P =+=

Evaluate γ: 30.1

39.341.4

V

P ===RR

ccγ


( )( )( ) ( )( ) kJ20.313.1

50.21L/mol22.4mol682.0kPa101.3 1.3

on =−

−−=W

96 •• Picture the Problem Let the subscripts 1 and 2 refer to the initial and final states respectively. Because the gas is initially at STP, we know that V1 = 22.4 L, P1 = 1 atm, and T1 = 273 K. We can use ( )12ln VVnRTW −= to find the work done on the gas


using 1

2211

−−

=γ


monatomic gas, γ = 1.67. (a) Express the work done on the gas in compressing it isothermally:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

1

2on ln

VVnRTW

Find the number of moles in 30 g of Ar (M = 40 g/mol):

mol750.0g/mol04

g30==n



1431


⎜⎝⎛⋅−=W


1

1

21

211

2211on

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−−

−=

γ

γ

VPPVP

VPVPW


γγ2211 VPVP = ⇒

γ

⎟⎟⎠

⎞⎜⎜⎝

⎛=

2

1

1

2

VV

PP

Substitute for P2/P1 and simplify to obtain:

1

2.01

1

5

12

111

1

2

1112

2

111

on −

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=γγγ

γγγ

VVVPV

VVVPV

VVVP

W


( )( )( ) ( )( ) kJ93.4167.1

50.21L/mol22.4mol75.0kPa101.3 1.67

on =−

−−=W

97 •• Picture the Problem We can use conservation of energy to relate the final temperature to the heat capacities of the gas and the solid. We can apply the Dulong-Petit law to find the heat capacity of the solid at constant volume and use the fact that the gas is diatomic to find its heat capacity at constant volume. Apply conservation of energy to this process:

0=∆Q

or ( ) ( ) 0K200K100 fsolidV,fgasV, =−−− TCTC

Solve for Tf: ( )( ) ( )( )

solidV,gasV,

solidV,gasV,f

K200K100CC

CCT

++

=

Chapter 18

1432

Using the Dulong-Petit law, determine the heat capacity of the solid at constant volume:

( )( )J/K49.9

KJ/mol8.314mol233solidV,

=⋅=

= nRC

Determine the heat capacity of the gas at constant volume:

( )( )J/K20.8

KJ/mol8.314mol125

25

gasV,

=

⋅=

= nRC


( )( ) ( )( ) K171J/K9.49J/K8.20

J/K9.49K200J/K8.20K100f =

++

=T

*98 •• Picture the Problem We can express the work done during an isobaric process as the product of the temperature and the change in volume and relate Q to ∆T through the definition of CP. Finally, we can use the first law of thermodynamics to show that ∆Eint = Cv∆T.

(a) . andonly of

function a is ly,Consequent . toalproportion is which molecules, gas theof energies kinetic theof sum theisenergy internal thegas, idealan For

Vint TCETUkT

∆=∆

(b) Use the first law of thermodynamics to relate the work done on the gas, the heat entering the gas, and the change in the internal energy of the gas:

oninint WQE +=∆

At constant pressure: ( ) ( ) TnRTTnRVVPW ∆=−=−= ififgasby

and TnRWW ∆−=−= gasby on

Relate Qin to CP and ∆T: TCQ ∆= Pin


( ) TCTnRC

TnRTCE

∆=∆−=

∆−∆=∆

VP

Pint


1433

99 •• Picture the Problem We can use TnRTCQ ∆=∆= 2

3Vin to find Qin for the constant-

volume process and TnRTCQ ∆=∆= 25

Pin to find Qin for the isobaric process. The

work done by the gas is given by .f

i

∫=V

V

PdVW Finally, we can apply the first law of

thermodynamics to find the change in the internal energy of the gas from the work done on the gas and the heat that enters the gas. (a) The heat added to the gas during this process is given by:

TnRQ ∆= 23

in


( )( )( )kJ3.74


in

=

⋅=Q

For a constant-volume process: 0gas by the =W

Apply the 1st law of thermodynamics to obtain:

oninint WQE +=∆ (1)

Substitute for Qin and Won in equation (1) and evaluate ∆Eint:

kJ74.30kJ74.3int =+=∆E

(b) Relate the heat absorbed by the gas to the change in its temperature:

( )( )( )kJ6.24


25

Pin

=

⋅=∆=∆= TnRTCQ

For a constant-pressure process that begins at temperature Ti and ends at temperature Tf, the work done by the gas is given by:

( )

( )if

ifgas by the

f

i

TTnR

VVPPdVWV

V

−=

−== ∫

Substitute numerical values and evaluate Wby the gas:

( )( )( )kJ49.2

K300KJ/mol314.8mol1gas by the

=

⋅=W

Apply the 1st law of thermodynamics to express the change in the internal energy of the gas during this isobaric expansion:

inonint QWE +=∆

Chapter 18

1434

Because the Wby gas = −Won: ingas by theint QWE +−=∆


kJ75.3kJ24.6 kJ49.2int =+−=∆E

Remarks: Because ∆Eint depends only on the initial and final temperatures of the gas, the values for ∆Eint for Part (a) and Part (b) should be they same. They differ slightly due to rounding. *100 •• Picture the Problem We can use Qin = CP∆T to find the change in temperature during this isobaric process and the first law of thermodynamics to relate W, Q, and ∆Eint. We can use TnRE ∆=∆ 2

5int to find the change in the internal energy of the gas during the

isobaric process and the ideal-gas law for a fixed amount of gas to express the ratio of the final and initial volumes. (a) Relate the change in temperature to Qin and CP and evaluate ∆T:

( )( )K8.59

KJ/mol8.314mol2J500

27

27

in

P

in

=

⋅=

==∆nR

QCQT

(b) Apply the first law of thermodynamics to relate the work done on the gas to the heat supplied and the change in its internal energy:

in25

inVininton

QTnRQTCQEW

−∆=−∆=−∆=


( )( )( )

J143J500

K8.59KJ/mol8.314mol225

on

−=−

⋅=W

Because Wby gas = −Won: J143gasby =W

(c) Using the ideal-gas law for a fixed amount of gas, relate the initial and final pressures, volumes and temperatures:

f

ff

i

ii

TVP

TVP

=

or, because the process is isobaric,

f

f

i

i

TV

TV

=


1435

Solve for and evaluate Vf/Vi:

03.1K293.15

K8.59K293.15i

i

i

f

i

f

=+

=

∆+==

TTT

TT

VV

101 •• Picture the Problem Knowing the rate at which energy is supplied, we can obtain the data we need to plot this graph by finding the time required to warm the ice to 0°C, melt the ice, warm the water formed from the ice to 100°C, vaporize the water, and warm the water to 110°C. Find the time required to warm the ice to 0°C:

( )( )( )

s20.0J/s100

K10KkJ/kg2kg0.1

ice1

=

⋅=

∆=∆

PTmct

Find the time required to melt the ice: ( )( )

s5.333J/s100

kJ/kg333.5kg0.1f2

=

==∆P

mLt

Find the time required to heat the water to 100°C:

( )( )( )

s418J/s100

K100KkJ/kg18.4kg0.1

w3

=

⋅=

∆=∆

PTmct

Find the time required to vaporize the water:

( )( )

s2257J/s100

kJ/kg2572kg0.1V4

=

==∆P

mLt

Find the time required to heat the vapor to 110°C:

( )( )( )

s20J/s100

K10KkJ/kg2kg0.1

steam5

=

⋅=

∆=∆

PTmct

Chapter 18

1436

The temperature T as a function of time t is shown to the right:

*102 •• Picture the Problem We know that, for an adiabatic process, Qin = 0. Hence the work done by the expanding gas equals the change in its internal energy. Because we’re given the work done by the gas during the expansion, we can express the change in the temperature of the gas in terms of this work and CV. Express the final temperature of the gas as a result of its expansion:

TTT ∆+= if

Apply the equation for adiabatic work and solve for ∆T:

TCW ∆−= Vadiabatic

and

nRW

CWT

25adiabatic

V

adiabatic −=−=∆

Substitute and evaluate Tf:

( )( )K216

KJ/mol8.314mol2kJ3.5K300

25

25adiabatic

if

=

⋅−=

−=nR

WTT

103 •• Picture the Problem Because PfVf = 4PiVi and Vf = Vi/2, the path for which the work done by the gas is a minimum while the pressure never falls below Pi is shown on the adjacent PV diagram. We can apply the first law of thermodynamics to relate the heat transferred to the gas to its change in internal energy and the work done on the gas.


1437

Using the first law of thermodynamics, relate the heat transferred to the gas to its change in internal energy and the work done on the gas:

oninint WQE +=∆

Solve for Qin: onintin WEQ −∆=

Express the work done during this process: ( )

RTnRTVPVVPVP

WWW

21

21

ii21

ii21

ii

volumeconstantprocess isobaricon

0===

−=+∆=

+=

because n = 1 mol.

Express ∆Eint for the process: ( )RT

TnRTnRTCE

29

23

23

Vint 3=

=∆=∆=∆

because n = 1 mol.

Substitute to obtain: RTRTRTQ 421

29

in =−=

104 •• Picture the Problem We can solve the ideal-gas law for the dilute solution for the increase in pressure and find the number of solute molecules dissolved in the water from their mass and molecular weight. Solve the ideal gas law for P to obtain:

VNkTP =

Express the number of solute molecules N in terms of the number of moles n and Avogadro’s number and then express the number of moles in terms of the mass of the salt and its molecular mass:

NaCl

AA M

mNnNN ==

Substitute to obtain: VM

kTmNPNaCl

A=


( )( )( )( )( )( )

2633

2323

N/m1027.1m10g/mol4.58

K297J/K10381.1molparticles/10022.6g30×=

××= −

−

P

Chapter 18

1438

105 •• Picture the Problem Let the subscripts 1 and 2 refer to the initial and final states in this adiabatic expansion. We can use an equation describing a quasi-static adiabatic process to express the final temperature as a function of the initial temperature and the initial and final volumes. Using the equation for a quasi-static adiabatic process, relate the initial and final volumes and temperatures:

111

122

−− = γγ VTVT

Solve for and evaluate T2: ( )( )

K396

2K300 14.11

2

112

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛= −

−γ

VVTT

106 •• Picture the Problem We can simplify our calculations by relating Avogadro’s number NA, Boltzmann’s constant k, the number of moles n, and the number of molecules N in the gas and solving for NAk. We can then calculate U300 K and U600 K and their difference. Express the increase in internal energy per mole resulting from the heating of diamond:

K300K600 UUU −=∆

Express the relationship between Avogadro’s number NA, Boltzmann’s constant k, the number of moles n, and the number of molecules N in the gas:

NknR = ⇒ kNknNR A==

Substitute in the given equation to obtain:

13

E

E

−= TTe

RTU

Determine U300 K: ( )( )

J7951

K1060KJ/mol314.83K300K1060K300

=−

⋅=

eU

Determine U600 K: ( )( )

kJ45.51

K1060KJ/mol314.83K600K1060K600

=−

⋅=

eU


1439


kJ4.66

J795kJ5.45K300K600

=

−=−=∆ UUU

*107 ••• Picture the Problem The isothermal expansion followed by an adiabatic compression is shown on the PV diagram. The path 1→2 is isothermal and the path 2→3 is adiabatic. We can apply the ideal-gas law for a fixed amount of gas and an isothermal process to find the pressure at point 2 and the pressure-volume relationship for a quasi-static adiabatic process to determineγ.

(a) Relate the initial and final pressures and volumes for the isothermal expansion and solve for and evaluate the final pressure:

2211 VPVP =

and

021

1

10

2

112 2

PV

VPVVPP ===

(b) Relate the initial and final pressures and volumes for the adiabatic compression:

γγ3322 VPVP =

or ( ) γγ

000021 32.12 VPVP =

which simplifies to 64.22 =γ

Take the natural logarithm of both sides of this equation and solve for and evaluate γ :

64.2ln2ln =γ

and

40.12ln64.2ln

==γ

diatomic. is gas the∴

(c) unchanged. is

energy kinetic onal translati theand constant, is process, isothermal In the T

1.32. offactor aby increasesenergy kinetic onal translati theand ,1.32 process, adiabatic In the 03 TT =

Chapter 18

1440

108 ••• Picture the Problem In this problem the specific heat of the combustion products depends on the temperature. Although CP increases gradually from (9/2)R per mol to (15/2)R per mol at high temperatures, we’ll assume that CP = 4.5R below T = 2000 K and CP = 7.5R above T = 2000 K. We’ll also use R = 2.0 cal/mol⋅K. We can find the final temperature following combustion from the heat made available during the combustion and the final pressure by applying the ideal-gas law to the initial and final states of the gases. (a) Relate the heat available in this combustion process to the change in temperature of the triatomic gases:

( )( )if

Pavailable

5.7 TTRnTnCQ

−=∆=

Solve for Tf to obtain: i

availablef 5.7

TnR

QT += (1)

Express Q available to heat the gases above 2000 K:

steamCOheat

K2000tomol9releasedavailable

2QQ

QQQ

−−

−= (2)

Express the energy released in the combustion of 1 mol of benzene:

( ) kcal758kcal151621

released ==Q

Noting that there are 3 mol of H2O and 6 mol of CO2, find the heat required to form the products at 100°C:

( )( )( )( )( )( )( )

kcal33.10cal/g540g/mol18mol3

K300373KKcal/g1g/mol18mol3

Vwwwsteam

=+

−⋅×=

+∆= LnMTcnMQ

and

( )( )( )

kcal942.3300KK373

Kcal/mol2mol65.4

5.4PCOheat 2

=−×

⋅=

∆=∆= TnRTnCQ

Find Q required to heat 9 mol of gas to 2000 K: ( )( )

( )kcal93.43

373KK0002Kcal/mol2mol95.4

5.4PK2000tomol9

=−×

⋅=

∆=∆= TnRTnCQ


1441


kcal589.2kcal33.10kcal3.94

kcal131.79kcal758available

=−−

−=Q

Substitute in equation (1) and evaluate Tf: ( )( )

K6364

K0002Kcal/mol2mol97.5

kcal589.2f

=

+⋅

=T

Apply the ideal-gas law to express the final volume in terms of the final temperature and pressure:

( )( )( )

3

f

ff

m70.4

kPa101.3K6364KJ/mol8.314mol9

=

⋅=

=P

nRTV

(b) Apply the ideal-gas law to relate the final temperature, pressure, and volume to the number of moles in the final state:

ffff RTnVP =

Apply the ideal-gas law to relate the initial temperature, pressure, and volume to the number of moles in the initial state:

iiii RTnVP =

Divide the first of these equations by the second and solve for Pf: ii

ff

ii

ff

RTnRTn

VPVP

=

or, because Tf = Ti,

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

f

i

i

fif V

VnnPP (3)

Find the initial volume Vi occupied by 8.5 mol of gas at 300 K and 1 atm:

( )( )

L209.2K273K300mol8.5L/mol22.4i

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=V

Chapter 18

1442

Substitute numerical values in equation (3) and evaluate Vf:

( )

atm0471.0

kPa101.325atm1kPa774.4

L4700L209.2

mol8.5mol9kPa101.3f

=

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=P

*109 ••• Picture the Problem In this problem the specific heat of the combustion products depends on the temperature. Although CP increases gradually from (9/2)R per mol to (15/2)R per mol at high temperatures, we’ll assume that CP = 4.5R below T = 2000 K and CP = 7.5R above T = 2000 K. We can find the final temperature following combustion from the heat made available during the combustion and the final pressure by applying the ideal-gas law to the initial and final states of the gases. (a) Apply the ideal-gas law to find the pressure due to 3 mol at 300 K in the container prior to the reaction:

( )( )( )

kPa5.93

L80K300KJ/mol8.314mol3

i

ii

=

⋅=

=V

nRTP

(b) Relate the heat available in this adiabatic process to CV and the change in temperature of the gases:

( )ifV

availableint

TTCQE

−==∆

Because T > 2000 K: ( ) nRnRRnnRCC 5.65.7PV =−=−=

Substitute to obtain: ( )ifavailable 5.6 TTnRQ −=

Solve for Tf to obtain: i

availablef 5.6

TnR

QT += (1)

Find Q required to raise 2 mol of CO2 to 2000 K:

TCQ ∆= VCOheat 2

For T < 2000 K: ( ) nRnRRnnRCC 5.35.4PV =−=−=


1443

Substitute for CV and find the heat required to warm to CO2 to 2000 K: ( )( )

( )kJ94.98

300KK0002KJ/mol.3148mol25.3

5.32COheat

=−×

⋅=

∆= TnRQ

Find Q available to heat 2 mol of CO2 above 2000 K: kJ1.461

kJ94.98kJ560available

=−=Q

Substitute in equation (1) and evaluate Tf:

( )( )K6266

K2000KJ/mol8.314mol26.5

kJ461.1f

=

+⋅

=T

Apply the ideal-gas law to relate the final temperature, pressure, and volume to the number of moles in the final state:

ffff RTnVP =

Apply the ideal-gas law to relate the initial temperature, pressure, and volume to the number of moles in the initial state:

iiii RTnVP =

Divide the first of these equations by the second and solve for Pf: ii

ff

ii

ff

RTnRTn

VPVP

=

or, because Vf = Vi,

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

i

f

i

fif T

TnnPP (2)

Substitute numerical values in equation (2) and evaluate Pf:

( )

MPa30.1

K300K6266

mol3mol2kPa53.39f

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=P

(c) Substitute numerical values in equation (2) and evaluate Pf for Tf = 273 K:

( )

kPa7.56

K300K273

mol3mol2kPa53.39f

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=P

Chapter 18

1444

110 ••• Picture the Problem The molar heat capacity at constant volume is related to the internal

energy per mole according todTdU

nc' 1

V = . We can differentiate U with respect to

temperature and use nR = Nk or R = NAk to establish the result given in the problem statement. From Problem 106 we have, for the internal energy per mol: 1

3E

EA

−= TTe

kTNU

Relate the molar heat capacity at constant volume to the internal energy per mol:

dTdU

nc' 1

V =

Use dTdU

nc' 1

V = to express :V'c

( ) ( )

( ) ( )22

22E

2EEEA

V

13

113

11

131

131

31

E

EE

E

E

EEE

−⎟⎠⎞

⎜⎝⎛=⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−

−=

−⎥⎥⎦

⎤

⎢⎢⎣

⎡

−

−=⎥⎦

⎤⎢⎣⎡

−=⎥⎦

⎤⎢⎣⎡

−=

TT

TTEETT

TT

TT

TTTTTT'

ee

TTR

TTe

eRT

edTd

eRT

edTdRT

ekTN

dTd

nc

111 ••• Picture the Problem We can rewrite our expression for 'cV by dividing its numerator and

denominator by TTe E and then using the power series for ex to show that, for T > TE, Rc' 3V ≈ . In part (b), we can use the result of Problem 103 to obtain values for 'cV every

100 K between 300 K and 600 K and use this data to find ∆U numerically. (a) From Problem 110 we have:

( )2

2E

V1

3E

E

−⎟⎠⎞

⎜⎝⎛=

TT

TT'

ee

TTRc

Divide the numerator and denominator by TTe E to obtain:

TTTT

TT

TTTT'

eeTTR

eeeT

TRc

EE

E

EE

213

1213

2E

2

2E

V

−+−⎟⎠⎞

⎜⎝⎛=

+−⎟⎠⎞

⎜⎝⎛=


1445

Apply the power series expansion to obtain:

E

2E

2EE

2EE

for

...2112...

2112 EE

TTTT

TT

TT

TT

TTee TTTT

>⎟⎠⎞

⎜⎝⎛≈

+⎟⎠⎞

⎜⎝⎛+−+−+

⎟⎠⎞

⎜⎝⎛++=+− −

Substitute to obtain: R

TTT

TRc' 313 2E

2E

V =

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛≈

(b) Use the result of Problem 110 to verify the table to the right:

T cV

(K) (J/mol⋅K) 300 9.65 400 14.33 500 17.38 600 19.35

The following graph of specific heat as a function of temperature shown to the right was plotted using a spreadsheet program:

5

7

9

11

13

15

17

19

21

300 350 400 450 500 550 600

T (K)

CV (J

/mol

-K)

Chapter 18

1446

Integrate numerically, using the formula for the area of a trapezoid, to obtain:

( )( ) ( )( )( )( )

,kJ62.4

KJ/mol35.1938.17K100KJ/mol38.1733.14K100KJ/mol33.1465.9K100

21

21

21

=

⋅++⋅++⋅+=∆U

a result in good agreement (< 1% difference) with the result of Problem 106. 112 ••• Picture the Problem In (a) we’ll assume that τ = f (A/V, T, k, m) with the factors dependent on constants a, b, c, and d that we’ll find using dimensional analysis. In (b) we’ll use our result from (a) and assume that the diameter of the puncture is about 2 mm, that the tire volume is 0.1 m3, and that the air temperature is 20°C. (a) Express τ = f (A/V, T, k, m):

( ) ( ) ( ) dcba

mkTVA⎟⎠⎞

⎜⎝⎛=τ (1)

Rewrite this equation in terms of the dimensions of the physical quantities to obtain:

( ) ( ) ( ) dc

ba MKT

MLKLT 2

21

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

where K represents the dimension of temperature.

Simplify this dimensional equation to obtain:

dcccba MTKLMKLT -2c21 −−= or

2c21 TMKLT −+−−= dccbac

Equate exponents to obtain: 12:T =− c , 02:L =− ac ,

0:K =− cb ,

and 0:M =+ dc


21−=c ,

1−=a ,

21−=b ,

and 21=d


1447

Substitute in equation (1): ( ) ( ) ( )

kTm

AV

mkTVA

=

⎟⎠⎞

⎜⎝⎛= −−

−

21

21

21

1

τ

(b) Substitute numerical values and evaluate τ:

( )( )( )

( )( ) min3.87s232K293KJ/mol314.8

m1.0kg/m293.1

m1024

m1.0 33

23

3

==⋅×

=−πτ

Chapter 18

1448

1449

Chapter 19 The Second Law of Thermodynamics Conceptual Problems 1 • Determine the Concept Friction reduces the efficiency of the engine. *2 • Determine the Concept As described by the second law of thermodynamics, more heat must be transmitted to the outside world than is removed by a refrigerator or air conditioner. The heating coils on a refrigerator are inside the room–the refrigerator actually heats the room it is in. The heating coils on an air conditioner are outside, so the waste heat is vented to the outside. 3 • Determine the Concept Increasing the temperature of the steam increases the Carnot efficiency, and generally increases the efficiency of any heat engine. 4 •• Determine the Concept To condense, water must lose heat. Because its entropy change is given by dS = dQrev/T and dQrev is negative, the entropy of the water decreases.

correct. is )(c

*5 • Determine the Concept (a) Because the temperature changes during an adiabatic process, the internal energy of the system changes continuously during the process. (b) Both the pressure and volume change during an adiabatic process and hence work is done by the system. (c) ∆Q = 0 during an adiabatic process. Therefore ∆S = 0. correct. is )(c

(d) Because the pressure and volume change during an adiabatic process, so does the temperature. 6 •• (a) False. The complete conversion of mechanical energy into heat is not prohibited by either the 1st or 2nd laws of thermodynamics and is common place in energy transformations. (b) True. This is the heat-engine statement of the 2nd law of thermodynamics.

Chapter 19

1450

(c) False. The efficiency of a heat engine is a function of the thermodynamic processes of its cycle. (d) False. With the input of sufficient energy, a heat pump can transfer a given quantity of heat from a cold reservoir to a hot reservoir. (e) False. The only restriction that the refrigerator statement of the 2nd law places on the COP is that it can not be infinite. (f) True. The Carnot engine, as a consequence of its thermodynamic processes, is reversible. (g) False. The entropy of one system can decrease at the expense of one or more other systems. (h) True. This is one statement of the 2nd law of thermodynamics. 7 •• Determine the Concept The two paths are shown on the PV diagram to the right. We can use the concept of a state function to choose from among the alternatives given as possible answers to the problem.

(a) Because Eint is a state function and the initial and final states are the same for the two paths and B int,A int, EE ∆=∆ .

(b) and (c) S, like Eint, is a state function and its change when the system moves from one state to another depends only on the system’s initial and final states. It is not dependent on the process by which the change occurs and BA SS ∆=∆ .

(d) correct. is )(d

The Second Law of Thermodynamics

1451

*8 •• Determine the Concept The processes A→B and C→D are adiabatic; the processes B→C and D→A are isothermal. The cycle is therefore the Carnot cycle shown in the adjacent PV diagram.

9 •• Determine the Concept Note that A→B is an adiabatic expansion. B→C is a constant volume process in which the entropy decreases; therefore heat is released. C→D is an adiabatic compression. D→A is a constant volume process that returns the gas to its original state. The cycle is that of the Otto engine (see Figure 19-3). 10 •• Determine the Concept Refer to Figure 19-3. Here a→b is an adiabatic compression, so S is constant and T increases. Between b and c, heat is added to the system and both S and T increase. c→d is again isentropic, i.e., without change in entropy. d→a releases heat and both S and T decrease. The cycle on an ST diagram is sketched in the adjacent figure.

11 •• Determine the Concept Referring to Figure 19-8, process 1→2 is an isothermal expansion. In this process heat is added to the system and the entropy and volume increase. Process 2→3 is adiabatic, so S is constant as V increases. Process 3→4 is an isothermal compression in which S decreases and V also decreases. Finally, process 4→1 is adiabatic, i.e., isentropic, and S is constant while V decreases. The cycle is shown in the adjacent SV diagram.

Chapter 19

1452

12 •• Picture the Problem The SV diagram of the Otto cycle is shown in Figure 19-13. (see Problem 9)

13 •• Determine the Concept Process A→B is at constant entropy, i.e., an adiabatic process in which the pressure increases. Process B→C is one in which P is constant and S decreases; heat is exhausted from the system and the volume decreases. Process C→D is an adiabatic compression. Process D→A returns the system to its original state at constant pressure. The cycle is shown in the adjacent PV diagram.

*14 • Picture the Problem Let ∆T be the change in temperature and ε = (Th − Tc)/Th be the initial efficiency. We can express the efficiencies of the Carnot engine resulting from the given changes in temperature and examine their ratio to decide which has the greater effect on increasing the efficiency.

If Th is increased by ∆T , ε′, the new efficiency is:

TTTTT'

∆+−∆+

=h

chε

If Tc is reduced by ∆T, the efficiency is:

h

ch

TTTT'' ∆+−

=ε


1h

h

h

ch

h

ch

>∆+

=

∆+−∆+

∆+−

=T

TT

TTTTT

TTTT

'''εε

reservoir.hot theof re temperatuin the increase equalan than more efficiency theincreases

by reservoir cold theof re temperatuin thereduction a Therefore, T∆


1453

Estimation and Approximation 15 •• Picture the Problem The maximum efficiency of an automobile engine is given by the efficiency of a Carnot engine operating between the same two temperatures. We can use the expression for the Carnot efficiency and the equation relating V and T for a quasi-static adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio. Express the Carnot efficiency of an engine operating between the temperatures Tc and Th:

h

cC 1

TT

−=ε

Relate the temperatures Tc and Th to the volumes Vc and Vh for a quasi-static adiabatic compression from Vc to Vh:

1hh

1cc

−− = γγ VTVT

Solve for the ratio of Tc to Th:

1

c

h1

c

1h

h

c

−

−

−

⎟⎟⎠

⎞⎜⎜⎝

⎛==

γ

γ

γ

VV

VV

TT


1

c

hC 1

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

γ

εVV

Express the compression ratio r:

h

c

VVr =

Substitute once more to obtain:

1C11 −−= γε

r

Substitute numerical values for r and γ (1.4 for diatomic gases) and evaluate εC:

( )%5.56565.0

811 14.1C ==−= −ε

*16 •• Picture the Problem If we assume that the temperature on the inside of the refrigerator is 0°C (273 K) and the room temperature to be about 30°C (303 K), then the refrigerator must be able to maintain a temperature difference of about 30 K. We can use the definition of the COP of a refrigerator and the relationship between the temperatures of the hot and cold reservoir and hQ and Qc to find an upper limit on the COP of a household refrigerator. In (b) we can solve the definition of COP for Qc and differentiate the resulting equation with respect to time to estimate the rate at which heat is being drawn from the refrigerator compartment. (a) Using its definition, express the COP of a household refrigerator: W

QcCOP = (1)

Chapter 19

1454

Apply the 1st law of thermodynamics to the refrigerator to obtain:

hc QQW =+

Substitute for W and simplify to obtain:

1

1COP

c

hch

c

−

=−

=

QQQQ

Q

Assume, for the sake of finding the upper limit on the COP, that the refrigerator is a Carnot refrigerator and relate the temperatures of the hot and cold reservoirs to hQ and Qc:

c

h

c

h

TT

QQ

=


1

1COP

c

hmax

−=

TT

Substitute numerical values and evaluate COPmax: 10.9

1K273K3031COPmax =−

=

(b) Solve equation (1) for Qc: ( )COPc WQ = (2)

Differentiate equation (2) with respect to time to obtain:

( )dt

dWdt

dQ COPc =

Substitute numerical values and evaluate dQc/dt: ( )( ) kW46.5J/s6009.10c ==

dtdQ

17 •• Picture the Problem We can use the definition of intensity to find the total power of sunlight hitting the earth and the definition of the change in entropy to find the changes in the entropy of the earth and the sun resulting from the radiation from the sun. (a) Using its definition, express the intensity of the sun’s radiation on the earth in terms of the power delivered to the earth P and the earth’s cross sectional area A:

API =

Solve for P and substitute for A to obtain:

2RIIAP π== where R is the radius of the earth.


1455

Substitute numerical values and evaluate P: ( )( )

W1066.1

m1037.6kW/m3.117

262

×=

×= πP

(b) Express dSearth/dt for the earth due to the flow of solar radiation:

earth

earth

TP

dtdS

=

Substitute numerical values and evaluate dSearth/dt:

sJ/K1072.5

K290W1066.1

14

17

earth

⋅×=

×=∆S

(c) Express dSsun/dt for the sun due to the outflow of solar radiation hitting the earth:

sun

sun

TP

dtdS

=

Substitute numerical values and evaluate dSsun/dt:

sJ/K1007.3

K5400W1066.1

13

17sun

⋅×=

×=

dtdS

18 •• Picture the Problem We can use the definition of intensity to find the total power radiated by the sun and the definition of the change in entropy to find the change in the entropy of the universe resulting from the radiation of 1011 stars in 1011 galaxies. (a) Using its definition, express the intensity of the sun’s radiation on the location of earth in terms of the total power it delivers to space P and the area of a sphere A whose radius is the distance from the sun to the earth:

API =

Solve for P and substitute for A to obtain:

24 IRIAP π== where R is the distance from the sun to the earth.

Substitute numerical values and evaluate P: ( )( )

W1068.3

m105.1kW/m3.1426

2112

×=

×= πP

(b) Express ∆Suniverse:

universeuniverse T

PS =∆

Chapter 19

1456

Substitute numerical values and evaluate ∆Suniverse:

( )

sJ/K1035.1

K73.2W1068.310

48

2622

universe

⋅×=

×=∆S

19 •• Picture the Problem We can use the definition of entropy change to estimate the increase in entropy of the universe as a result of the heat produced by a typical human body. The entropy change is equivalent to the entropy change if the heat from the body were added to the universe reversibly. Express the increase in entropy of the universe as a result of the heat produced by a human body:

night

night

day

dayu T

QTQ

S∆

+∆

=∆

Using the definition of power, express the total heat produced by a human body:

tPQ ∆=∆

Assume that half of the heat is produced during the day and half at night:

tPQQ ∆=∆=∆ 21

nightday


⎟⎟⎠

⎞⎜⎜⎝

⎛+∆=

∆+

∆=∆

nightday21

night

21

day

21

u

11TT

tP

TtP

TtPS

Use ( ) 27332F9

5 +−= tT to obtain: Tday = 294 K and Tnight = 286 K

Substitute numerical values and evaluate ∆Su:

( )( )( ) kJ/K8.29K286

1K294

1s/h3600h/d24J/s10021

u =⎟⎟⎠

⎞⎜⎜⎝

⎛+=∆S

*20 ••• Picture the Problem If you had one molecule in a box, it would have a 50% chance of being on one side or the other. We don’t care which side the molecules are on as long as they all are on one side, so with one molecule you have a 100% chance of it being on one side or the other. With two molecules, there are four possible combinations (both on one side, both on the other, one on one side and one on the other, and the reverse), so there is a 25% (1 in 4) chance of them both being on a particular side, or a 50% chance of them both being on either side. Extending this logic, the probability of N molecules all being on one side of the box is P = 2/2N, which means that, if the molecules shuffle 100 times a second, the time it would take them to cover all the combinations and all get on one side

or the other is ( )10022N

t = . In (e) we can apply the ideal gas law to find the number of


1457

molecules in 1 L of air at a pressure of 10−12 torr and an assumed temperature of 300 K. (a) Evaluate t for N = 10 molecules: ( ) s12.5

1002210

==t

(b) Evaluate t for N = 100 molecules: ( )

y1001.2

s1034.610022

20

27100

×=

×==t

(c) Evaluate t for N = 1000 molecules: ( )1002

21000

=t

To evaluate 10002 let 1000210 =x and take the logarithm of both sides of the equation to obtain:

( ) 10ln2ln1000 x=


301=x


y1058.1

s105.01002

10

290

299301

×=

×==t

(d) Evaluate t for N = 6.02×1023 molecules: ( )1002

2231002.6 ×

=t

To evaluate

231002.62 × let 231002.6210 ×=x and take the

logarithm of both sides of the equation to obtain:

( ) 10ln2ln1002.6 23 x=×


2310≈x


( ) y101002

10 2323

1010

≈≈t

(e) Solve the ideal gas law for the number of molecules N in the gas: kT

PVN =

Assuming the gas to be at room temperature (300 K), substitute numerical values and evaluate N:

( )( )( )( )( )

molecules1022.3K300J/K10381.1

m10Pa/torr32.133torr10

7

23

3312

×=

×= −

−−

N

Chapter 19

1458

Evaluate T for N = 3.22×107 molecules: ( )1002

271022.3 ×

=t

To evaluate

71022.32 × let 71022.3210 ×=x and take the

logarithm of both sides of the equation to obtain:

( ) 10ln2ln1022.3 7 x=×


710≈x

Substitute to obtain: ( ) y10

100210 7

7

1010

≈=T

Express the ratio of this waiting time to the lifetime of the universe Tuniverse:

77

1010

10

universe

10y10y10≈=

TT

or

universe107

10 TT ≈

Heat Engines and Refrigerators 21 • Picture the Problem We can use the definition of the efficiency of a heat engine to relate the work done W, the heat absorbed Qin, and the heat rejected each cycle Qout.

(a) Express Qin in terms of W and ε :

J5000.2

J100in ===

εWQ

(b) Solve the definition of efficiency for and evaluate outQ :

( ) ( )( )J400

2.01J5001inout

=

−=−= εQQ

22 • Picture the Problem We can use its definition to find the efficiency of a heat engine from the work done, the heat absorbed, and the heat rejected each cycle.

(a) Use the definition of the efficiency of a heat engine:

%30J400J120

in

==≡QWε

(b) Solve the definition of efficiency for and evaluate outQ :

( ) ( )( )J280

3.01J4001inout

=

−=−= εQQ


1459

23 • Picture the Problem We can use its definition to find the efficiency of the engine and the definition of power to find its power output.

(a) Apply the definition of the efficiency of a heat engine:

%0.40J100J601

Q1

in

out =−=−=Qε

(b) Use the definition of power to find the power output of this engine:

( ) W80.0s0.5

J1000.4in ==∆

=∆∆

=t

Qt

WP ε

*24 • Picture the Problem We can apply their definitions to find the COP of the refrigerator and the efficiency of the heat engine.

(a) Using the definition of the COP, relate the heat absorbed from the cold reservoir to the work done each cycle:

WQcCOP =

Relate the work done per cycle to Qh and Qc:

ch QQW −=

Substitute to obtain: ch

cCOPQQ

Q−

=

Substitute numerical values and evaluate COP:

67.1kJ5kJ8

kJ5COP =−

=

(b) Use the definition of efficiency to relate the work done per cycle to the heat absorbed from the high-temperature reservoir:

hQW

=ε

Substitute numerical values and evaluate ε :

%5.37kJ8kJ3

==ε

Chapter 19

1460

25 •• Picture the Problem To find the heat added during each step we need to find the temperatures in states 1, 2, 3, and 4. We can then find the work done on or by the gas along each pass from the area under each straight-line segment and the heat that enters or leaves the system from TCQ ∆= V and .P TCQ ∆= We can find the efficiency of the

cycle from the work done each cycle and the heat that enters the system each cycle.

(a) The cycle is shown to the right:

Apply the ideal-gas law to state 1 to find T1:

( )( )( )( )


L24.6atm12

111

=⋅⋅×

=

=

−

nRVPT

The pressure doubles while the volume remains constant between states 1 and 2. Hence:

KTT 6002 12 ==

The volume doubles while the pressure remains constant between states 2 and 3. Hence:

KTT 12002 23 ==

The pressure is halved while the volume remains constant between states 3 and 4. Hence:

KTT 600321

4 ==

For path 1→2:

02121 =∆= →→ VPW

and ( )( )

kJ74.3

K300K600KJ/mol8.31423

2123

21V21int,21

=

−⋅=∆=∆=∆= →→→→ TRTCEQ


1461

For path 2→3:

( )( )

kJ99.4atmL

J101.325atmL20.49L24.6L49.2atm22232

=⋅

×⋅=−=∆= →→ VPW

and ( )( )

kJ5.12

K600K1200KJ/mol8.31425

3225

32P32

=

−⋅=∆=∆= →→→ TRTCQ

For path 3→4:

04343 =∆= →→ VPW

and ( )( )

kJ48.7

K0021K600KJ/mol8.31423

4323

43V43int,43

−=

−⋅=∆=∆=∆= →→→→ TRTCEQ

For path 4→1:

( )( )

kJ49.2atmL

J101.3atmL6.24L2.94L24.6atm11414

=⋅

×⋅−=−=∆= →→ VPW

and ( )( )

kJ24.6

K600K003KJ/mol8.31425

1425

14P14

−=

−⋅=∆=∆= →→→ TRTCQ

(b) Use its definition to find the efficiency of this cycle:

%4.15kJ12.5kJ3.74kJ2.49kJ4.99

3221

1432

in

=+−

=

++

==→→

→→

QQWW

QWε

Remarks: Note that the work done per cycle is the area bounded by the rectangular path.

Chapter 19

1462

26 •• Picture the Problem The three steps in the process are shown on the PV diagram. We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.

Express the efficiency of the cycle: inQ

W=ε

Find the heat entering or leaving the system during the adiabatic expansion:

01 =Q

Find the heat entering or leaving the system during the isobaric compression:

( )( ) Latm35L20L10atm127

227

227

2V2

⋅−=−=

∆=∆=∆= VPTRTCQ

Find the heat entering or leaving the system during the constant-volume process:

( )( )Latm41

L10atm1-atm2.6425

325

325

3V3

⋅==

∆=∆=∆= PVTRTCQ

Apply the 1st law of thermodynamics to the cycle ( 0cycle int, =∆E ) to

obtain:

Latm6Latm41L35atm0

321

inininton

⋅=⋅+⋅−=

++=−=−∆=

QQQQQEW

Substitute and evaluate ε : %6.14

Latm41Latm6

=⋅⋅

=ε

27 •• Picture the Problem We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.


1463

Apply the ideal-gas law to states 1, 2, 3, and 4 to find the pressures at these points:

( )( )( ) atm33.1K24.6


1

11 =

⋅⋅×==

−

VnRTP

Proceed as above to obtain the values shown in the table:

Point P V T

(atm) (L) (K) 1 1.330 24.6 400 2 0.667 49.2 400 3 0.500 49.2 300 4 1.000 24.6 300

The PV diagram is shown to the right:

Express the efficiency of the cycle: inQ

W=ε (1)

Find the work done by the gas and the heat that enters the system during the isothermal expansion from 1 to 2:

( )( )( )


1

212121

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛== →→ V

VnRTQW

Find the work done by the gas and the heat that enters the system during the constant-volume compression from 2 to 3:

032 =→W

and ( )( ) kJ2.10K400K300J/K2132V32 −=−=∆= →→ TCQ

Chapter 19

1464

Find the work done by the gas and the heat that enters the system during the isothermal expansion from 3 to 4:

( )( )( )


3

434343

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛== →→ V

VnRTQW

Find the work done by the gas and the heat that enters the system during the constant-volume process from 4 to 1:

014 =→W

and ( )( ) kJ2.10K300K400J/K2114V14 =−=∆= →→ TCQ

Evaluate the work done each cycle:

kJ0.57600kJ1.7290kJ2.305

14433221

=+−+=

+++= →→→→ WWWWW

Find the heat that enters the system each cycle: kJ4.405

kJ2.100kJ2.3051421in

=+=

+= →→ QQQ

Substitute numerical values in equation (1) and evaluate ε :

%1.13kJ4.405kJ0.5760

==ε

*28 •• Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas and the heat capacities at constant volume and constant pressure to find the heat flow for the constant-volume and isobaric processes. Because the change in internal energy is zero for the isothermal process, we can use the expression for the work done on or by a gas during an isothermal process to find the heat flow during such a process. Finally, we can find the efficiency of the cycle from its definition.

(a) Use the ideal-gas law to find the temperature at point 1:

( )( )( )( )

K301

KJ/mol8.314mol1L25kPa10011

1

=

⋅==

nRVPT


1465

Use the ideal-gas law to find the temperatures at points 2 and 3:

( )( )( )( )

K601

KJ/mol8.314mol1L25kPa20022

32

=

⋅===

nRVPTT

(b) Find the heat entering or leaving the system for the constant-volume process from 1 to 2:

( )( )kJ3.74

K301K601KJ/mol8.31423

2123

21V21

=

−⋅=∆=∆= →→→ TRTCQ

Find the heat entering or leaving the system for the isothermal process from 2 to 3:

( )( )( ) kJ46.3L52L05lnK016KJ/mol8.314mol1ln

2

3232 =⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=→ V

VnRTQ

Find the heat entering or leaving the system during the isobaric compression from 3 to 1:

( )( )

kJ24.6

K601K301KJ/mol8.31425

1325

13P13

−=

−⋅=∆=∆= →→→ TRTCQ

(c) Express the efficiency of the cycle: 3221in →→ +

==QQ

WQWε (1)

Apply the 1st law of thermodynamics to the cycle: kJ0.960

kJ6.24kJ3.46kJ.743133221

=−+=

++== →→→∑ QQQQW

because, for the cycle, ∆U = 0.


%3.13kJ3.46kJ3.74

kJ0.960=

+=ε

29 •• Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas. We can find the efficiency of the cycle from its definition; using the area enclosed by the cycle to find the work done per cycle and the heat entering the system between states 1 and 2 and 2 and 3 to determine Qin.

Chapter 19

1466

(a) Use the ideal-gas law for a fixed amount of gas to find the temperature in state 2 to the temperature in state 1:

2

22

1

11

TVP

TVP

=

Solve for and evaluate T2:

( ) ( )( )

K600

atm1atm3K200

1

21

11

2212

=

===PPT

VPVPTT

Apply the ideal-gas law for a fixed amount of gas to states 2 and 3 to obtain:

( ) ( )( )

K1800

L100L300K600

2

32

22

3323

=

===VVT

VPVPTT

Apply the ideal-gas law for a fixed amount of gas to states 3 and 4 to obtain:

( ) ( )( )

K600

atm3atm1K1800

3

43

33

4434

=

===PPT

VPVPTT

(b) Express the efficiency of the cycle:

inQW

=ε (1)

Use the area of the rectangle to find the work done each cycle:

( )( )Latm400

atm1atm3L100L300⋅=

−−=∆∆= VPW

Apply the ideal-gas law to state 1 to find the product of n and R:

( )( )

atm/KL0.5K200

L100atm1

1

11

⋅=

==TVPnR

Noting that heat enters the system between states 1 and 2 and states 2 and 3, express Qin:

( )nRTTTnRTnR

TCTCQQQ

3227

2125

3227

2125

32P21V

3221in

→→

→→

→→

→→

∆+∆=

∆+∆=

∆+∆=+=


( )[ ( )]( ) Latm2600atm/KL5.0K600K1800K200K600 27

25

in ⋅=⋅−+−=Q


1467


%4.15Latm2600Latm400

=⋅⋅

=ε

30 ••• Picture the Problem To find the efficiency of the diesel cycle we can find the heat that enters the system and the heat that leaves the system and use the expression that gives the efficiency in terms of these quantities. Note that no heat enters or leaves the system during the adiabatic processes a→b and c→d. Express the efficiency of the cycle in terms of Qc and Qh: h

c1QQ

−=ε

Express Q for the isobaric warming process b→c:

( )bccb TTCQQ −==→ Ph

Express Q for the constant-volume cooling process d→a:

( )adad TTCQQ −==→ Vc


( )( )bc

ad

bc

ad

TTTT

TTCTTC

−−

−=

−−

−=

γ

ε

1

1P

V

Using the equation of state for an adiabatic process, relate the temperatures Ta and Tb:

11 −− = γγbbaa VTVT (1)

Proceeding similarly, relate the temperatures Tc and Td:

11 −− = γγddcc VTVT (2)

Use equations (1) and (2) to eliminate Ta and Td: ( )

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

−=

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−−

−

−

−

−

c

b

a

b

c

b

a

c

bc

a

bb

d

cc

TT

VV

TT

VV

TTVVT

VVT

11

1

11

1

1

1

1

γ

γε

γγ

γ

γ

γ

γ

because Va = Vd.

Chapter 19

1468

Apply the ideal-gas law for a fixed amount of gas to relate Tb and Tc: c

b

c

b

VV

TT

=

because Pb = Pc.


⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

−=⋅

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

−=

−−−

a

b

a

c

a

b

a

c

a

b

a

c

a

b

a

b

a

c

a

c

a

c

c

b

a

b

c

b

a

c

VV

VV

VV

VV

VV

VV

VV

VV

VV

VVVV

VV

VV

VV

VV

γ

γγε

γγ

γγγγ

1

11

1

111

*31 •• Picture the Problem We can use the efficiency of a Carnot engine operating between reservoirs at body temperature and typical outdoor temperatures to find an upper limit on the efficiency of an engine operating between these temperatures. (a) Express the maximum efficiency of an engine operating between body temperature and 70°F:

h

cC 1

TT

−=ε

Use ( ) 27332F95 +−= tT to obtain: Tbody = 310 K and Troom = 294 K

Substitute numerical values and evaluate Cε :

%16.5K310K2941C =−=ε

need.t weenergy tha get the tofoodeat weRather, t.environmen theandbody ourbetween swappingheat fromenergy our get t don' that notecan we

but text in the discussedbeen not haveenergy body with supply thethat ones theassuch reactions chemical tolaw second theofn applicatio The

amics. thermodynof law second thecontradictnot doesbody human a ofefficiency actual than thelessly considerab is efficiency thisfact that The


1469

(b)

level.high ly unreasonaban at maintained be tohave wouldraturesbody tempe internal

,efficiency eappreciabl work withengineheat a make To humans. as conditions same eroughly thunder survive animals bloodedMost warm −

32 ••• Picture the Problem The Carnot cycle’s four segments (shown to the right) are: (A) an isothermal expansion at T = Th from V1 to V2, (B) an adiabatic expansion from V2 to V3, (C) an isothermal compression from V3 to V4 at T = Tc, and (D) an adiabatic compression from V4 to V1. We can find the Carnot efficiency for a gas described by the Clausius equation by expressing the ratio of the work done per cycle to the heat entering the system per cycle.

Express the efficiency of the Carnot cycle in terms of the work done and the heat that enters the system per cycle:

hQW

=ε

Apply the first law of thermodynamics to segment A:

h

1

2hh

AA int,AA

ln2

1

2

1

Q

bnVbnVnRT

bnVdVnRT

PdVWEWQ

V

V

V

V

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=−

=

==∆+=

∫

∫

Follow the same procedure for segment C to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=−

=

==∆+=

∫

∫

bnVbnVnRT

bnVdVnRT

PdVWEWQ

V

V

V

V

3

4cc

CC int,CC

ln4

3

4

3

and

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=bnVbnVnRTQ

4

3cc ln

Chapter 19

1470

Apply the first law of thermodynamics to the complete cycle ( 0cycle int, =∆E ) to express W:

⎠

⎞⎜⎜⎝

⎛−−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=

−=

bnVbnVnRT

bnVbnVnRT

QQW

4

3c

1

2h

ch

lnln


⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=

bnVbnVT

bnVbnVT

bnVbnVnRT

bnVbnVnRT

bnVbnVnRT

1

2h

4

3c

1

2h

4

3c

1

2h

ln

ln1

ln

lnlnε

Apply the first law of thermodynamics to the adiabatic processes B and D:

dTCdVbnV

nRTdTCPdVdEdWdQ

V

VB int,BB 0

+−

=

+=+==

Separate variables and integrate to obtain:

( )∫

∫∫

−−−=

−−=

bnVdVbnV

dVCnT

TdT

1

V

γ

or ( ) ( )( ) constantln

constantln1ln1 +−=

+−−−=−γ

γ

bnV

bnVT

Simplify to obtain: ( ) constantlnln 1 =−+ −γbnVT

or ( ) constantln 1 =− −γbnVT

and ( ) constant1 =− −γbnVT

Using this result, relate V2 and V3 to Th and Tc:

( ) ( ) 13c

12h

−− −=− γγ bnVTbnVT (1)

Relate V1 and V4 to Th and Tc:

( ) ( ) 14c

11h

−− −=− γγ bnVTbnVT (2)

Divide equation (1) by equation (2) and simplify to obtain:

( )( )

( )( ) 1

4h

13c

11h

12h

−

−

−

−

−−

=−−

γ

γ

γ

γ

bnVTbnVT

bnVTbnVT


1471

or

bnVbnV

bnVbnV

−−

=−−

4

3

1

2

Substitute in our expression for ε and simplify:

h

c

1

2h

1

2c

1ln

ln1

TT

bnVbnVT

bnVbnVT

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

−=ε

the same as for an ideal gas.

Second Law of Thermodynamics 33 •• Determine the Concept The relationship of the perfect engine and the refrigerator to each other and to the hot and cold reservoirs is shown below. To remove 500 J from the cold reservoir and reject 800 J to the hot reservoir, 300 J of work must be done on the system. Assuming that the heat-engine statement is false, one could use the 800 J rejected to the hot reservoir to do 300 J of work. Thus, running the refrigerator connected to the ″perfect″ heat engine would have the effect of transferring 500 J of heat from the cold to the hot reservoir without any work being done, in violation of the refrigerator statement of the second law.

*34 •• Determine the Concept The work done by the system is the area enclosed by the cycle, where we assume that we start with the isothermal expansion. It is only in this expansion that heat is extracted from a reservoir. There is no heat transfer in the adiabatic expansion or compression. Thus, we would completely convert heat to mechanical energy, without exhausting any heat to a cold reservoir, in violation of the second law.

Chapter 19

1472

Carnot Engines 35 • Picture the Problem We can find the efficiency of the Carnot engine using

hc /1 TT−=ε and the work done per cycle from ./ hQW=ε We can apply conservation

of energy to find the heat rejected each cycle from the heat absorbed and the work done each cycle. We can find the COP of the engine working as a refrigerator from its definition.

(a) Express the efficiency of the Carnot engine in terms of the temperatures of the hot and cold reservoirs:

%3.33K300K20011

h

cC =−=−=

TTε

(b) Using the definition of efficiency, relate the work done each cycle to the heat absorbed from the hot reservoir:

( )( ) J33.3J1000.333hC === QW ε

(c) Apply conservation of energy to relate the heat given off each cycle to the heat absorbed and the work done:

J66.7J33.3J100hc =−=−= WQQ

(d) Using its definition, express and evaluate the refrigerator’s coefficient of performance:

00.2J33.3J66.7COP c ===

WQ

36 • Picture the Problem We can find the efficiency of the engine from its definition and the additional work done if the engine were reversible from ,hCQW ε= where εC is the

Carnot efficiency.

(a) Express the efficiency of the engine in terms of the heat absorbed from the high-temperature reservoir and the heat exhausted to the low-temperature reservoir:

%0.20J250J2001

1h

c

h

ch

h

=−=

−=−

==QQ

QQQ

QWε

(b) Express the additional work done if the engine is reversible:

aWWW partCarnot −=∆


1473

Relate the work done by a reversible engine to its Carnot efficiency:

( ) J3.83J250K300K2001

1 hh

chC

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−== Q

TTQW ε

Substitute and evaluate ∆W: J33.3J50J3.38 =−=∆W

37 •• Determine the Concept Let the first engine be run as a refrigerator. Then it will remove 140 J from the cold reservoir, deliver 200 J to the hot reservoir, and require 60 J of energy to operate. Now take the second engine and run it between the same reservoirs, and let it eject 140 J into the cold reservoir, thus replacing the heat removed by the refrigerator. If ε2, the efficiency of this engine, is greater than 30%, then Qh2, the heat removed from the hot reservoir by this engine, is 140 J/(1 − ε2) > 200 J, and the work done by this engine is W = ε2Qh2 > 60 J. The end result of all this is that the second engine can run the refrigerator, replacing the heat taken from the cold reservoir, and do additional mechanical work. The two systems working together then convert heat into mechanical energy without rejecting any heat to a cold reservoir, in violation of the second law.

38 •• Determine the Concept If the reversible engine is run as a refrigerator, it will require 100 J of mechanical energy to take 400 J of heat from the cold reservoir and deliver 500 J to the hot reservoir. Now let the second engine, with ε2 > 0.2, operate between the same two heat reservoirs and use it to drive the refrigerator. Because ε2 > 0.2, this engine will remove less than 500 J from the hot reservoir in the process of doing 100 J of work. The net result is then that no net work is done by the two systems working together, but a finite amount of heat is transferred from the cold to the hot reservoir, in violation of the refrigerator statement of the second law.

*39 •• Picture the Problem We can use the definition of efficiency to find the efficiency of the Carnot engine operating between the two reservoirs. (a) Use its definition to find the efficiency of the Carnot engine:

%3.33J150J50

hC ===

QWε

(b) If COP > 2, then 50 J of work will remove more than 100 J of heat from the cold reservoir and put more than 150 J of heat into the hot reservoir. So running engine (a) to operate the refrigerator with a COP > 2 will result in the transfer of heat from the cold to the hot reservoir without doing any net mechanical work in violation of the second law.

Chapter 19

1474

40 •• Picture the Problem We can use the definitions of the efficiency of a Carnot engine and the coefficient of performance of a refrigerator to find these quantities. The work done each cycle by the Carnot engine is given by hCQW ε= and we can use the conservation

of energy to find the heat rejected to the low-temperature reservoir.

(a) Use the definition of the efficiency of a Carnot engine to obtain:

74.3%K300K7711

h

cC =−=−=

TTε

(b) Express the work done each cycle in terms of the efficiency of the engine and the heat absorbed from the high-temperature reservoir:

( )( ) J3.74J100743.0hC === QW ε

(c) Apply conservation of energy to obtain:

J25.7J74.3J001hc =−=−= WQQ

(d) Using its definition, express and evaluate the refrigerator’s coefficient of performance:

346.0J74.3J25.7COP c ===

WQ

41 •• Picture the Problem We can use the ideal-gas law for a fixed amount of gas and the equations of state for an adiabatic process to find the temperatures, volumes, and pressures at the end points of each process in the given cycle. We can use

TQ ∆= VC and TQ ∆= PC to find the heat entering and leaving during the constant-

volume and isobaric processes and the first law of thermodynamics to find the work done each cycle. Once we’ve calculated these quantities, we can use its definition to find the efficiency of the cycle and the definition of the Carnot efficiency to find the efficiency of a Carnot engine operating between the extreme temperatures.

(a) Apply the ideal-gas law for a fixed amount of gas to relate the temperature at point 3 to the temperature at point 1:

3

33

1

11

TVP

TVP

=


1

313 VVTT = (1)

Apply the ideal-gas law for a fixed amount of gas to relate the pressure at point 2 to the temperatures at points 1 and 2 and the pressure at 1:

2

22

1

11

TVP

TVP

=

or, because V1 = V2,


1475

( ) atm1.55K273K423atm1

1

212 ===

TTPP

Apply the state equation for an adiabatic process to relate the pressures and volumes at points 2 and 3:

γγ3311 VPVP =

Noting that V1 = 22.4 L, solve for and evaluate V3: ( )

L30.6atm1

atm1.55L22.41.411

3

113

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ

PPVV

Substitute in equation (1) and evaluate T3:

( ) K373L22.4L30.6K2733 ==T

and C10027333 °=−= Tt

(b) Process 1→2 takes place at constant volume (note that γ = 1.4 corresponds to a diatomic gas and that CP – CV = R):

( )( )kJ3.12

K273K423KJ/mol8.314C

25

2125

21V21

=

−⋅=

∆=∆= →→→ TRTQ

Process 2→3 takes place adiabatically:

032 =→Q

Process 3→1 is isobaric (note that CP = CV + R):

( )( )kJ2.91

K373K732KJ/mol8.314C

27

2127

13P13

−=

−⋅=

∆=∆= →→→ TRTQ

(c) Use its definition to express the efficiency of this cycle:

inQW

=ε

Apply the first law of thermodynamics to the cycle:

oninint WQK +=∆ or, because 0cycle int, =∆E (the system begins

and ends in the same state) and Won = − Wby the gas = W, inQW = .

Chapter 19

1476

Evaluate W:

kJ0.210kJ2.910kJ3.12

133221

=−+=

++== →→→∑ QQQQW

Substitute and evaluate ε : %73.6

kJ3.12kJ0.210

==ε

(d) Express and evaluate the efficiency of a Carnot cycle operating between 423 K and 273 K:

35.5%K234K73211

h

cC =−=−=

TTε

42 •• Picture the Problem We can find the maximum efficiency of the steam engine by calculating the Carnot efficiency of an engine operating between the given temperatures. We can apply the definition of efficiency to find the heat discharged to the engine’s surroundings in 1 h.

(a) Find the efficiency of a Carnot engine operating between these temperatures:

%5.40K543K32311

h

cmax =−=−=

TTε

Find the efficiency of the steam engine as a percentage of the maximum possible efficiency:

maxmaxenginesteam 741.0405.030.0 εεε ==

(b) Relate the heat discharged to the engine’s surroundings to Qh and the efficiency of the engine:

( ) hc 1 QQ ε−=

Using its definition, relate the efficiency of the engine to the heat intake of the engine and the work it does each cycle:

εεtPWQ ∆

==h

Substitute and evaluate cQ in 1 h: ( )

( ) ( )( )

GJ68.10.3

s3600kJ/s2003.01

1c

=

−=

∆−=

εε tPQ


1477

Heat Pumps *43 • Picture the Problem We can use the definition of the COPHP and the Carnot efficiency of an engine to express the maximum efficiency of the refrigerator in terms of the reservoir temperatures. We can apply equation 19-10 and the definition of power to find the minimum power needed to run the heat pump.

(a) Express the COPHP in terms of Th and Tc:

ch

h

h

c

h

c

h

hhHP

1

1

1

1

COP

TTT

TT

QQ

QQQ

WQ

c

−=

−=

−=

−==

Substitute numerical values and evaluate the COPHP:

26.6K263K313

K133COPHP =−

=

(b) Using its definition, express the power output of the engine:

tWP∆

=

Use equation 19-10 to express the work done by the heat pump:

HP

h

COP1+=

QW

Substitute and evaluate P: kW75.2

6.261kW20

COP1 HP

h =+

=+

∆=

tQP

(c) Find the minimum power if the COP is 60% of the efficiency of an ideal pump:

( ) ( )kW21.4

6.266.01kW20

COP6.01 maxHP,

cmin

=

+=

+∆

=tQ

P

44 • Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc, Th, P, and ∆t.

Chapter 19

1478

(a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP:

( )( ) tP

WQ∆=

=COPCOPc

Express the COP in terms of Th and Tc:

ch

c

h

c

h

h

h

cc

11

1111

COP

TTT

TT

QWQ

QQ

WQ

−=

−−

=−=−

=

−===

εεε

εε


tPTT

TQ ∆⎟⎟⎠

⎞⎜⎜⎝

⎛−

=ch

cc

Substitute numerical values and evaluate Qc:

( )( )

kJ303

s60W370K273K293

K273c

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=Q

(b) Find the heat removed if the COP is 70% of the efficiency of an ideal pump:

( ) ( )( )

kJ122

s60W370K273K293

K2737.0c

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

='Q

45 • Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc, Th, P, and ∆t.

(a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP:

( )( ) tP

WQ∆=

=COPCOPc


1479

Express the COP in terms of Th and Tc:

ch

c

h

c

h

h

h

cc

11

1

111

COP

TTT

TT

QWQ

QQ

WQ

−=−

−=

−=−

=

−===

εεε

εε


tPTT

TQ ∆⎟⎟⎠

⎞⎜⎜⎝

⎛−

=ch

cc

Substitute numerical values and evaluate Qc:

( )( )

kJ731

s60W370K273K083

K273c

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=Q

(b) Find the heat removed if the COP is 70% of the efficiency of an ideal pump:

( ) ( )( )

kJ121

s60W370K273K083

K2737.0c

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

='Q

Entropy Changes 46 • Picture the Problem We can use the definition of entropy change to find the change in entropy of the water as it freezes.

Apply the definition of entropy change to obtain: T

mLTQS f−=

∆=∆

Substitute numerical values and evaluate ∆S:

( )( ) J/K22.0K273

J/g333.5g18−=

−=∆S

*47 •• Picture the Problem The change in the entropy of the world resulting from the freezing of this water and the cooling of the ice formed is the sum of the entropy changes of the water-ice and the freezer. Note that, while the entropy of the water decreases, the entropy of the freezer increases. Express the change in entropy of the universe resulting from this freezing and cooling process:

freezerwateru SSS ∆+∆=∆ (1)

Chapter 19

1480

Express ∆Swater:

coolingfreezingwater SSS ∆+∆=∆ (2)

Express ∆Sfreezing:

freezing

freezingfreezing T

QS

−=∆ (3)

where the minus sign is a consequence of the fact that heat is leaving the water as it freezes.

Relate Qfreezing to the latent heat of fusion and the mass of the water:

ffreezing mLQ =

Substitute in equation (3) to obtain: freezing

ffreezing T

mLS −=∆

Express ∆Scooling:

⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

i

fpcooling ln

TTmCS

Substitute in equation (2) to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+

−=∆

i

fp

freezing

fwater ln

TTmC

TmLS

Noting that the freezer gains heat (at 263 K) from the freezing water and cooling ice, express ∆Sfreezer:

freezer

p

freezer

f

freezer

ice cooling

freezer

icefreezer

TTmC

TmL

TQ

TQS

∆+=

∆+

∆=∆

Substitute for ∆Swater and ∆Sfreezer in equation (1):

⎥⎥⎦

⎤

⎢⎢⎣

⎡ ∆++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

−=

∆++⎟⎟

⎠

⎞⎜⎜⎝

⎛+

−=∆

freezer

p

freezer

f

i

fp

freezing

f

freezer

p

freezer

f

i

fp

freezing

fu

ln

ln

TTC

TL

TTC

TLm

TTmC

TmL

TTmC

TmLS


1481


( ) ( )

( )( )

J/K40.2

K263K263K273KJ/kg2100

K263J/kg105.333

K273K263lnKJ/kg2100

K273J/kg105.333kg05.0

33

u

=

⎥⎦

⎤−⋅+

⎢⎣

⎡ ×+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅+

×−=∆S

and, because ∆Su > 0, the entropy of the universe increases. 48 • Picture the Problem We can use the definition of entropy change and the first law of thermodynamics to express ∆S for the ideal gas as a function of its initial and final volumes.

(a) Use its definition to express the entropy change of the gas:

TQS ∆

=∆

Apply the first law of thermodynamics to the isothermal process:

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−∆=∆

i

fonint ln

VVnRTWEQ



( )( )

J/K11.5


lni

f

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

VVnRS

(b) Because the process is reversible: 0u =∆S

Remarks: The entropy change of the environment of the gas is −11.5 J/K.

49 • Picture the Problem We can use the definition of entropy change and the 1st law of thermodynamics to express ∆S for the ideal gas as a function of its initial and final volumes.

Chapter 19

1482

(a) Use its definition to express the entropy change of the gas:

TQS ∆

=∆


⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−∆=∆

i

fonint ln

VVnRTWEQ



( )( )

J/K11.5


lni

f

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

VVnRS

(b) Because the process is not quasi-static, it is non-reversible:

0u >∆S

50 • Picture the Problem We can use the definition of entropy change to find the change in entropy of the water as it changes to steam.


mLTQS v=

∆=∆


( )( ) kJ/K06.6K373MJ/kg26.2kg1

==∆S

51 • Picture the Problem We can use the definition of entropy change to find the change in entropy of the ice as it melts.


mLTQS f=

∆=∆


( )( ) kJ/K22.1K273

kJ/kg5.333kg1==∆S

52 •• Picture the Problem We can use the first law of thermodynamics to find the change in the internal energy of the system and the change in the entropy of the system from the


1483

change in entropy of the hot- and cold-reservoirs.

(a) Apply the 1st law of thermodynamics to find the change in the internal energy of the system:

( )J50

J50J100J200oninint

=

−−=+=∆ WQE

(b) Express the change in entropy of the system as the sum of the entropy changes of the high- and low-temperature reservoirs:

J/K0.167K200J100

K300J200

c

c

h

hch

=−=

−=∆−∆=∆TQ

TQSSS

(c) Because the process is reversible: 0u =∆S

(d) Because Ssystem is a state function: J50int =∆E , J/K0.167=∆S ,

and 0u >∆S

*53 •• Picture the Problem We can use the fact that the system returns to its original state to find the entropy change for the complete cycle. Because the entropy change for the complete cycle is the sum of the entropy changes for each process, we can find the temperature T from the entropy changes during the 1st two processes and the heat rejected during the third.

(a) Because S is a state function of the system:

0cyclecomplete =∆S

(b) Relate the entropy change for the complete cycle to the entropy change for each process:

03

2

2

1

1 =++TQ

TQ

TQ


0J400K400J200

K300J300

=−

++T

Solve for T: K267=T

54 •• Picture the Problem We can use the definition of entropy change and the 1st law of

Chapter 19

1484

thermodynamics to express ∆S for the ideal gas as a function of its initial and final volumes. (a) Use its definition to express the entropy change of the gas:

TQS ∆

=∆


⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−∆=∆

i

fonintin ln

VVnRTWEQ

because ∆Eint = 0 for free expansion.


( )( )

J/K11.5


lni

f

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

VVnRS

(b) Because the process is irreversible, Su > 0 and, because no heat is exchanged:

J/K5.11u =∆S

55 •• Picture the Problem Because the ice gains heat as it melts, its entropy change is positive and can be calculated from its definition. Because the temperature of the lake is just slightly greater than 0°C and the mass of water is so much greater than that of the block of ice, the absolute value of the entropy change of the lake will be approximately equal to the entropy change of the ice as it melts.

(a) Use the definition of entropy change to find the entropy change of the ice:

( )( )

kJ/K244

K273kJ/kg333.5kg200f

ice

=

==∆T

mLS

(b) Relate the entropy change of the lake to the entropy change of the ice:

kJ/K244icelake −=∆−≈∆ SS

(c) Because the temperature of the lake is slightly greater than that of the ice, the magnitude of the entropy change of the lake is less than 244 kJ/K and the entropy change of the universe is greater than zero. The melting of the ice is an irreversible process and 0u >∆S .


1485

56 •• Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a melting process and for constant-pressure processes to find the entropy change of the universe, i.e., the piece of ice and the water in the insulated container.

(a) Apply conservation of energy to obtain:

gainedlost QQ =

or waterwarmingicemeltingwatercooling QQQ +=

Substitute to relate the masses of the ice and water to their temperatures, specific heats, and the final temperature of the water:

( )( )( ) ( )( ) ( )( )( )tt °⋅+=−°°⋅ Ccal/g1g100cal/g79.7g100C100Ccal/g1g100

Solve for t to obtain: C2.10 °=t

(b) Express the entropy change of the universe:

watericeu SSS ∆+∆=∆

Using the expression for the entropy change for a constant-pressure process, express the entropy change of the melting ice and warming ice-water:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

∆+∆=∆

i

fP

f

f

waterwarmingicemeltingice

lnTTmc

TmL

SSS


( )( ) ( )( ) J/K138K273K283.2lnKkJ/kg4.184kg0.1

K273kJ/kg333.5kg0.1

ice =⎟⎟⎠

⎞⎜⎜⎝

⎛⋅+=∆S

Find the entropy change of the cooling water:

( )( ) J/K115K373K283.2lnKkJ/kg4.18kg0.1water −=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=∆S

Substitute for ∆Sice and ∆Swater and evaluate the entropy change of the universe:

J/K23.0

J/K115J/K381u

=

−=∆S

Chapter 19

1486

Remarks: The result that ∆Su > 0 tells us that this process is irreversible. *57 •• Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a constant pressure process to find the entropy changes of the copper block, the water, and the universe.

(a) Using the equation for the entropy change during a constant-pressure process, express the entropy change of the copper block:

⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

i

fCuCuCu ln

TTcmS (1)


gainedlost QQ =

or waterwarmingblockcopper QQ =

Substitute to relate the masses of the block and water to their temperatures, specific heats, and the final temperature Tf of the water: ( )( )( ) ( )( )( )( )K15.273KkJ/kg 4.184kg/L1L4K15.373KkJ/kg0.386kg1 ff −⋅=−⋅ TT

Solve for Tf: K275.40 f =T

Substitute numerical values in equation (1) and evaluate ∆SCu:

( )( ) J/K117K373.15K275.40lnKkJ/kg0.386kg1Cu −=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=∆S

(b) Express the entropy change of the water:

⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

i

fwaterwaterwater ln

TTcmS

Substitute numerical values and evaluate ∆Swater:

( )( ) J/K137K273.15K40.275lnKkJ/kg184.4kg4water =⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=∆S


1487

(c) Substitute for ∆SCu and ∆Swater and evaluate the entropy change of the universe:

J/K20.3

J/K137J/K117waterCuu

=

+−=∆+∆=∆ SSS

Remarks: The result that ∆Su > 0 tells us that this process is irreversible. 58 •• Picture the Problem Because the mass of the water in the lake is so much greater than the mass of the piece of lead, the temperature of the lake will increase only slightly and we can reasonably assume that its final temperature is 10°C. We can apply the equation for the entropy change during a constant pressure process to find the entropy changes of the piece of lead, the water in the lake, and the universe.

Express the entropy change of the universe in terms of the entropy changes of the lead and the water in the lake:

wPbu SSS ∆+∆=∆

Using the equation for the entropy change during a constant-pressure process, express and evaluate the entropy change of the lead:

( )( ) J/K66.70K373.15K15.283lnKkJ/kg0.128kg2ln

i

fPbPbPb −=⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

TTcmS

Find the entropy change of the water in the lake:

( )( )( )

J/K37.81K283.15

K90KkJ/kg0.128kg2w

PbPbPb

w

Pb

w

ww

=

⋅=

∆===∆

TTcm

TQ

TQS

Substitute and evaluate ∆Su:

J/K10.7

J/K81.37J/K70.66u

=

+−=∆S

59 •• Picture the Problem Because the air temperature will not change appreciably as a result of this crash; we can assume that the kinetic energy of the car is transformed into heat at a temperature of 20°C. We can use the definition of entropy change to find the entropy change of the universe.

Chapter 19

1488

Express the entropy change of the universe as a consequence of the kinetic energy of the car being transformed into heat:

Tmv

TQS

221

u ==∆

Substitute numerical values and evaluate ∆Su: ( )

kJ/K1.97

K293.15s3600

h1h

km100kg15002

21

u

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×

=∆S

*60 •• Picture the Problem The total change in entropy resulting from the mixing of these gases is the sum of the changes in their entropies. (a) Express the total change in entropy resulting from the mixing of the gases:

BA SSS ∆+∆=∆

Express the change in entropy of each of the gases:

⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

iA

fAA ln

VVnRS

and

⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

iB

fBB ln

VVnRS

Because the initial and final volumes of the gases are the same and both volumes double:

( )2ln2ln2i

f nRVVnRS =⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆


( )( ) ( )J/K5.11

2lnKJ/mol314.8mol12

=

⋅=∆S

(b)mechanics. quantum

using derivedbeen has phonomenon thisofn descriptio completeA change.t doesn'entropy theishable,indistingu are molecules gas theBecause

Entropy and Work Lost *61 •• Picture the Problem We can find the entropy change of the universe from the entropy changes of the high- and low-temperature reservoirs. The maximum amount of the 500 J of heat that could be converted into work can be found from the maximum efficiency of an engine operating between the two reservoirs.


1489

(a) Express the entropy change of the universe:

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

+−=∆+∆=∆

ch

chchu

11TT

Q

TQ

TQSSS


( )

J/K0.417

K3001

K4001J500u

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=∆S

(b) Express the heat that could have been converted into work in terms of the maximum efficiency of an engine operating between the two reservoirs:

hmaxQW ε=

Express the maximum efficiency of an engine operating between the two reservoir temperatures:

h

cCmax 1

TT

−== εε

Substitute and evaluate W: ( )

J125

J500K400K30011 h

h

c

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−= Q

TTW

62 •• Picture the Problem Although in the adiabatic free expansion no heat is lost by the gas, the process is irreversible and the entropy of the gas increases. In the isothermal reversible process that returns the gas to its original state, the gas releases heat to the surroundings. However, because the process is reversible, the entropy change of the universe is zero. Consequently, the net entropy change is the negative of that of the gas in the isothermal compression. (a) Relate the entropy change of the universe to the entropy change of the gas during the isothermal compression:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=∆−=∆

i

fgasu ln

VVnRSS

Chapter 19

1490


( )( ) J/K5.76L6.24L3.12lnKJ/mol8.314mol1u =⎟⎟⎠

⎞⎜⎜⎝

⎛⋅−=∆S

(b) cycle.

in the wasted wasnone and done work wasno ,reversible and isothermal wasexpansion initial t theextent tha theTo

(c) Express the wasted work in terms of T and the entropy change of the universe:

( )( )kJ1.73

J/K5.76K300ulost

=

=∆= STW

General Problems 63 • Picture the Problem We can use the definition of power to find the work done each cycle and the definition of efficiency to find the heat that is absorbed each cycle. Application of the first law of thermodynamics will yield the heat given off each cycle.

(a) Use the definition of power to relate the work done in each cycle to the period of each cycle:

( )( )20.0J

s0.1W200cycle

=

=∆= tPW

(b) Express the heat absorbed in each cycle in terms of the work done and the efficiency of the engine:

J66.70.3

J20cyclecycleh, ===

εW

Q

Apply the 1st law of thermodynamics to find the heat given off in each cycle: J46.7

J20J66.7cycleh,cyclec,

=

−=−= WQQ

64 • Picture the Problem We can use their definitions to find the efficiency of the engine and that of a Carnot engine operating between the same reservoirs.

(a) Apply the definition of efficiency:

%7.16J150J12511

h

c

h

=−=−==QQ

QWε


1491

(b) Find the efficiency of a Carnot engine operating between the same reservoirs:

%4.2115.37315.29311

h

cC =−=−=

TTε

Express the ratio of the two efficiencies:

780.0%4.21%7.16

C

==εε

65 • Picture the Problem We can use the definition of efficiency to find the work done by the engine during each cycle and the first law of thermodynamics to find the heat exhausted in each cycle.

(a) Express the efficiency of the engine in terms of the efficiency of a Carnot engine working between the same reservoirs:

%0.51K500K200185.0

185.085.0h

cC

=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−==

TTεε

(b) Use the definition of efficiency to find the work done in each cycle:

( ) kJ102kJ200.510h === QW ε

(c) Apply the first law of thermodynamics to the cycle to obtain: kJ0.89

kJ021kJ002cycleh,cyclec,

=

−=−= WQQ

*66 •• Picture the Problem We can use the expression for the Carnot efficiency of the plant to find the highest efficiency this plant can have. We can then use this efficiency to find the power that must be supplied to the plant to generate 1 GW of power and, from this value, the power that is wasted. The rate at which heat is being delivered to the river is related to the requisite flow rate of the river by .dtdVTcdtdQ ρ∆=

(a) Express the Carnot efficiency of a plant operating between temperatures Tc and Th:

h

cCmax 1

TT

−== εε

Substitute numerical values and evaluate εC:

404.0K500K2981max =−=ε

(c) Find the power that must be supplied, at 40.4% efficiency, to produce an output of 1 GW:

GW48.20.404GW1

max

outputsupplied ===

εP

P

Chapter 19

1492

(b) Relate the wasted power to the power generated and the power supplied:

generatedsuppliedwasted PPP −=

Substitute numerical values and evaluate Pwasted:

GW48.1GW1GW48.2wasted =−=P

(d) Express the rate at which heat is being dumped into the river:

( )

dtdVTc

VdtdTc

dtdmTc

dtdQ

ρ

ρ

∆=

∆=∆=

Solve for the flow rate dV/dt of the river:

ρTcdtdQ

dtdV

∆=

Substitute numerical values (see Table 19-1 for the specific heat of water) and evaluate dV/dt:

( )( )( )L/s1008.7s/m708

kg/m10K5.0J/kg4180J/s1048.1

53

33

9

×==

×=

dtdV

67 • Picture the Problem We can find the rate at which the house contributes to the increase in the entropy of the universe from the ratio of ∆S to ∆t.

Using the definition of entropy change, express the rate of increase in the entropy of the universe:

TtQ

tTQ

tS ∆∆

=∆

∆=

∆∆

Substitute numerical values and evaluate ∆S/∆t:

W/K113K266

kW30==

∆∆

tS

68 •• Picture the Problem Because the cycle represented in Figure 19-12 is a Carnot cycle, its efficiency is that of a Carnot engine operating between the temperatures of its isotherms.

Express the Carnot efficiency of the cycle: h

cC 1

TT

−=ε

Substitute numerical values and evaluate εC:

%0.60K750K3001C =−=ε


1493

69 •• Picture the Problem All 500 J of mechanical energy are lost, i.e., transformed into heat in process (1). For process (2), we can find the heat that would be converted to work by a Carnot engine operating between the given temperatures and subtract amount of work from 1 kJ to find the energy that is lost. In part (b) we can use its definition to find the change in entropy for each process. (a) For process (2):

inCrecoveredmax,2 QWW ε==

Find the efficiency of a Carnot engine operating between 400 K and 300 K:

25.040030011

h

cC =−=−=

KK

TTε


( ) J250kJ10.25recovered ==W

or 750 J are lost.

energy. of wastefulmoreis (2) Process energy. of wastefulmore is (1) Process

totalmechanical

(b) Find the change in entropy of the universe for process (1):

J/K1.67K300J500

1 ==∆

=∆TQS

Express the change in entropy of the universe for process (2):

⎟⎟⎠

⎞⎜⎜⎝

⎛−∆=

∆+

∆−=∆+∆=∆

hc

chch2

11TT

Q

TQ

TQSSS

Substitute numerical values and evaluate ∆S2:

( )

J/K833.0

K4001

K3001kJ12

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=∆S

70 •• Picture the Problem Denote the three states of the gas as 1, 2, and 3 with 1 being the initial state. We can use the ideal-gas law and the equation of state for an adiabatic process to find the temperatures, volumes, and pressures at points 1, 2, and 3. To find the work done during each cycle, we can use the equations for the work done during isothermal, isobaric, and adiabatic processes. Finally, we find the efficiency of the cycle from the work done each cycle and the heat that enters the system during the isothermal expansion.

Chapter 19

1494

(a) Apply the ideal-gas law to the isothermal expansion 1→2 to find P2:

( ) atm4L4L1atm16

2

112 ===VVPP

Apply the equation of state for an adiabatic process to relate the pressures and volumes at 1 and 3:

γγ3311 VPVP =

and

( )

L2.29atm4atm16L1

1.6711

3

113

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ

PPVV

The PV diagram is shown to the right:

(b) From (a) we have:

L29.23 =V

Apply the equation of state for an adiabatic process (γ =1.67) to relate the temperatures and volumes at 1 and 3:

111

133

−− = γγ VTVT

and

( )

K344

L2.29L1K600

11.671

3

113

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−γ

VVTT

(c) Express the work done each cycle:

133221 →→→ ++= WWWW (1)

For the process 1→2:

( )( )

Latm22.2L1L4lnL1atm16

lnln1

211

1

2121

⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=→ V

VVPVVnRTW

For the process 2→3:

( )( )Latm84.6

L4L2.29atm432232

⋅−=−=

∆= →→ VPW


1495

For the process 3→1: ( )( )( )( ) ( )( )[ ]

Latm0.31L2.29atm4L1atm162

3

331123

3123

13V13

⋅−=

−−=

−−=

−−=∆−= →→

VPVPTTnRTCW

Substitute numerical values in equation (1) and evaluate W: Latm5.06

Latm10.3Latm6.84Latm2.22

⋅=

⋅−⋅−⋅=W

(d) Using its definition, express and evaluate the efficiency of the cycle: %8.22

Latm22.2Latm5.06

2121in

=⋅⋅

=

===→→ W

WQW

QWε

*71 •• Picture the Problem We can express the temperature of the cold reservoir as a function of the Carnot efficiency of an ideal engine and, given that the efficiency of the heat engine is half that of a Carnot engine, relate Tc to the work done by and the heat input to the real heat engine. Using its definition, relate the efficiency of a Carnot engine working between the same reservoirs to the temperature of the cold reservoir:

h

cC 1

TT

−=ε

Solve for Tc: ( )Chc 1 ε−= TT

Relate the efficiency of the heat engine to that of a Carnot engine working between the same temperatures:

C21

in

εε ==QW

orin

C2QW

=ε

Substitute to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

inhc

21QWTT

The work done by the gas in expanding the balloon is:

( )( ) Latm4L4atm1 ⋅==∆= VPW

Chapter 19

1496

Substitute numerical values and evaluate Tc:

( ) K313kJ4

LatmJ101.325Latm42

1K393.15c =

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛

⋅×⋅

−=T

72 •• Picture the Problem We can use the definitions of the COP and εC to show that their relationship is COP = Tc / (εCTh). Using the definition of the COP, relate the heat removed from the cold reservoir to the work done each cycle:

WQcCOP =

Apply energy conservation to relate Qc, Qh, and W:

WQQ −= hc

Substitute to obtain: W

WQ −= hCOP

Divide numerator and denominator by Qh and simplify to obtain:

h

hh

1COP

QW

QW

WWQ

−=

−=

Because hC QW=ε :

hC

c

C

h

c

C

h

c

C

C

111COP

TT

TT

TT

ε

εεεε

=

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=−

=

73 •• Picture the Problem We can use the definition of the COP to express the work the motor must do to maintain the temperature of the freezer in terms of the rate at which heat flows into the freezer. Differentiation of this expression with respect to time will yield an expression for the power of the motor that is needed to maintain the temperature in the freezer.


1497

Using the definition of the COP, relate the heat that must be removed from the freezer to the work done by the motor:

WQcCOP =

Solve for W: COP

cQW =

Differentiate this expression with respect to time to express the power of the motor:

COPc dtdQ

dtdWP ==

Express the maximum COP of the motor:

TT∆

= cmaxCOP


c

c

TT

dtdQP ∆

=

Substitute numerical values and evaluate P: ( ) W10.0

K250K50W50 =⎟⎟

⎠

⎞⎜⎜⎝

⎛=P

74 •• Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the isobaric, adiabatic, and isothermal processes of the cycle. (a) Apply the ideal-gas law to find the volume of the gas at A: ( )( )( )

L19.7atm

kPa101.325atm5


AA

=

×

⋅=

=P

nRTV

(b) We’re given that:

( ) L39.4L19.722 AB === VV

Apply the ideal-gas law to this isobaric process to obtain:

( ) K12002K600A

A

A

BAB ===

VV

VVTT

Chapter 19

1498

(c) Because the process C→A is isothermal:

K600AC == TT

(d) Apply the equation of state for an adiabatic process (γ = 1.4) to find the volume of the gas at C:

1CC

1BB

−− = γγ VTVT

and

( )

L223

K600K1200L39.4

11.41

11

C

BBC

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−γ

TTVV

(e) Express and evaluate the work done by the gas during the isobaric process AB:

( ) ( )( )( )

kJ9.98Latm

J101.325Latm98.50

L19.7atm52

AA

AAAABABA

=⋅

×⋅=

==−=−=−

VPVVPVVPW

Apply the first law of thermodynamics to express the work done by the gas during the adiabatic expansion BC:

CB25

CBVCB int,

CB int,CB in,CBint,CB on, 0

−

−−

−−−−

∆−=

∆−=∆=

−∆=−∆=

TnRTncE

EQEW

Substitute numerical values and evaluate WB−C:

( )( )( ) kJ24.9K1200K600KJ/mol8.314mol225

CB =−⋅−=−W

The work done by the gas during the isothermal compression CA is:

( )( )( ) kJ2.24L223L19.7lnK600KJ/mol8.314mol2ln

C

ACAC −=⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=− V

VnRTW

(f) The heat absorbed during the isobaric expansion AB is:

( )( )( )kJ9.34

K600K1200KJ/mol8.314mol227

BA27

BAPBA

=

−⋅=∆=∆= −−− TnRTncQ

The heat absorbed during the adiabatic expansion BC is:

0CB =−Q


1499

Use the first law of thermodynamics to find the heat absorbed during the isothermal compression CA:

kJ2.24ACAC int,ACAC

−=

=∆+= −−−− WEWQ

because ∆Eint,C−A = 0 for an isothermal process.

(g) The thermodynamic efficiency ε is: BA

ACCBBA

in −

−−− ++==

QWWW

QWε

Substitute numerical values and evaluate ε:

%6.30

kJ34.9kJ24.2kJ24.9kJ9.98

=

−+=ε

75 •• Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB:

( )

kPa253

atm1kPa101.325atm2.50

2atm5

A

A

B

AAB

=

×=

==V

VVVPP

(b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC:

BB

CCBC VP

VPTT =

Using the ideal-gas law, find the volume at B:

( )( )( )

L39.43kPa253

K600KJ/mol8.314mol2B

BB

=

⋅=

=P

nRTV

Use the equation of state for an adiabatic process and γ = 1.4 to find the volume occupied by the gas at C:

( )

L75.87atm1atm2.5L39.43

1.411

C

BBC

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ

PPVV

Chapter 19

1500

Substitute and evaluate TC: ( ) ( )( )( )( )

K462

L39.43atm2.5L75.87atm1K600C

=

=T

(c) Express the work done by the gas in one cycle:

ADDCCBBA −−−− +++= WWWWW

The work done during the isothermal expansion AB is:

( )( )( ) kJ6.915V

2VlnK600KJ/mol8.314mol2ln

A

A

A

BABA =⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=− V

VnRTW

The work done during the adiabatic expansion BC is:

( )( )( )kJ5.737


CB25

CBVCB

=−⋅−=∆−=∆−= −−− TnRTCW

The work done during the isobaric compression CD is:

( ) ( )( )

kJ5.690Latm

J101.325Latm56.17L75.87L19.7atm1CDCDC

−=⋅

×⋅−=−=−=− VVPW

Express and evaluate the work done during the constant-volume process DA:

0AD =−W

Substitute numerical values and evaluate W: kJ96.6

0kJ5.690kJ5.737kJ915.6

=

+−+=W

Using its definition, express the thermodynamic efficiency of the cycle:

ADBAin −− +==

QQW

QWε (1)

Express and evaluate the heat entering the system during the isothermal process AB:

kJ915.6BABA int,BABA ==∆+= −−−− WEWQ

Because ∆Eint = 0 for an isothermal process.

Express the heat entering the system AD25

ADVAD −−− ∆=∆= TnRTCQ


1501

during the constant-volume process DA: Apply the ideal-gas law to the constant-volume process DA to obtain:

( ) K120atm5atm1K600

A

DAD ===

PPTT

The heat entering the system during the process DA is:

( )( )( ) kJ0.20K120K600KJ/mol8.314mol225

AD =−⋅=−Q

Substitute numerical values in equation (1) and evaluate the efficiency of the cycle:

%9.25kJ20.0kJ6.915

kJ6.975=

+=ε

76 •• Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the isobaric, adiabatic, and isothermal processes of the cycle. (a) Apply the ideal-gas law to find the volume of the gas at A: ( )( )( )

L19.7atm

kPa101.325atm5


AA

=

×

⋅=

=P

nRTV

(b) We’re given that:

( ) L39.4L19.722 AB === VV

Apply the ideal-gas law to this isobaric process to obtain:

( ) K12002K600A

A

A

BAB ===

VV

VVTT

(c) Because the process CA is isothermal:

K600AC == TT

(d) Apply the equation of state for an adiabatic process (γ = 5/3) to find the volume of the gas at C:

1CC

1BB

−− = γγ VTVT

and

Chapter 19

1502

( )

L111

K600K1200L39.4

23

11

C

BBC

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−γ

TTVV

(e) Express and evaluate the work done by the gas during the isobaric process AB:

( ) ( )( )( )

kJ9.98Latm

J101.325Latm98.50

L19.7atm52

AA

AAAABABA

=⋅

×⋅=

==−=−=−

VPVVPVVPW

Apply the first law of thermodynamics to express the work done by the gas during the adiabatic expansion BC:

( )CB2

3

CBVCB int,

CB int,

CB in,CB int,CB on,

0

−

−−

−

−−−

∆−=

∆−=∆=

−∆=

−∆=

TnRTncE

EQEW

Substitute numerical values and evaluate WB−C:

( )( )( ) kJ14.9K1200K600KJ/mol8.314mol223

CB on, =−⋅−=−W

The work done by the gas during the isothermal compression CA is:

( )( )( )

kJ2.17

L111L19.7lnK600KJ/mol8.314mol2ln

C

ACAC

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=− V

VnRTW

(f) The heat absorbed during the isobaric expansion AB is:

( )( )( )kJ9.24

K600K1200KJ/mol8.314mol225

BA25

BAPBA in,

=

−⋅=∆=∆= −−− TnRTncQ

Express and evaluate the heat absorbed during the adiabatic expansion BC:

0CB =−Q


1503

Use the first law of thermodynamics to express and evaluate the heat absorbed during the isothermal compression CA:

kJ2.17ACAC int,ACAC

−=

=∆+= −−−− WEWQ


(g) The definition of thermodynamic efficiency ε is: BA

ACCBBA

in −

−−− ++==

QWWW

QWε

Substitute numerical values and evaluate ε:

%8.30

kJ24.9kJ.271kJ14.9kJ98.9

=

−+=ε

77 •• Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB:

( )

kPa253atm1

kPa101.3atm2.50

2atm5

A

A

B

AAB

=×=

==V

VVVPP

(b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC:

BB

CCBC VP

VPTT =

Using the ideal-gas law, find the volume at B:

( )( )( )

L39.43kPa253

K600KJ/mol8.314mol2B

BB

=

⋅=

=P

nRTV

Use the equation of state for an adiabatic process and γ = 5/3 to find the volume occupied by the gas at C:

( )

L33.86atm1atm2.5L39.43

531

C

BBC

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

γ

PPVV

Chapter 19

1504

Substitute and evaluate TC: ( ) ( )( )( )( )

K416

L39.43atm2.5L33.86atm1K600C

=

=T

(c) Express the work done by the gas in one cycle:

ADDCCBBA −−−− +++= WWWWW (1)

The work done during the isothermal expansion AB is:

( )( )( ) kJ6.9152lnK600KJ/mol8.314mol2lnA

A

A

BABA =⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=− V

VVVnRTW

The work done during the adiabatic expansion BC is:

( )( )( )kJ589.4


CB25

CBVCB

=−⋅−=∆−=∆−= −−− TnRTCW

The work done during the isobaric compression CD is:

( ) ( )( )

kJ926.4LatmJ101.3Latm63.84L33.86L19.7atm1CDCDC

−=⋅

×⋅−=−=−=− VVPW

The work done during the constant-volume process DA is:

0AD =−W

Substitute numerical values in equation (1) to obtain: kJ58.6

0kJ926.4kJ589.4kJ915.6

=

+−+=W

The thermodynamic efficiency of the cycle is given by:

ADBAin −− +==

QQW

QWε (2)

The heat entering the system during the isothermal process AB is:

kJ915.6BABA int,BABA

=

=∆+= −−−− WEWQ


The heat entering the system during the constant-volume process DA is:

AD23

ADVAD −−− ∆=∆= TnRTCQ


1505

Apply the ideal-gas law to the constant-volume process DA to obtain:

( ) K120atm5atm1K600

A

DAD ===

PPTT

The heat entering the system during the process DA is:

( )( )( ) kJ0.12K120K600KJ/mol8.314mol223

AD =−⋅=−Q

Substitute numerical values in equation (2) and evaluate the efficiency of the cycle:

%8.34kJ.021kJ6.915

kJ6.58=

+=ε

78 •• Picture the Problem We can express the efficiency of the Otto cycle using the result from Example 19-2. We can apply the relation constant1 =−γTV to the adiabatic processes of the Otto cycle to relate the end-point temperatures to the volumes occupied by the gas at these points and eliminate the temperatures at c and d. We can use the ideal-gas law to find the highest temperature of the gas during its cycle and use this temperature to express the efficiency of a Carnot engine. Finally, we can compare the efficiencies by examining their ratio. The efficiency of the Otto engine is given in Example 19-2:

b

ad

TTTT

−−

−=c

O 1ε (1)

where the subscripts refer to the various points of the cycle as shown in Figure 19-3.

Apply the relation constant1 =−γTV to the adiabatic

process a→b to obtain:

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γ

b

aab V

VTT

Apply the relation constant1 =−γTV to the adiabatic

process c→d to obtain:

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γ

c

ddc V

VTT

Subtract the first of these equations from the second to obtain:

11 −−

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛=−

γγ

b

aa

c

ddbc V

VTVVTTT

Chapter 19

1506

In the Otto cycle, Va = Vd and Vc = Vb. Substitute to obtain:

( )1

11

−

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛=−

γ

γγ

b

aad

b

aa

b

adbc

VVTT

VVT

VVTTT


( )

b

a

a

b

b

aad

ad

TT

VV

VVTT

TT

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−−=

−

−

11

1

1

1O

γ

γε

Note that, while Ta is the lowest temperature of the cycle, Tb is not the highest temperature.

Apply the ideal-gas law to c and b to obtain an expression for the cycle’s highest temperature Tc:

b

b

c

c

TP

TP

= ⇒ bb

cbc T

PPTT >=

Express the efficiency of a Carnot engine operating between the maximum and minimum temperatures of the Otto cycle:

c

a

TT

−=1Cε

Express the ratio of the efficiency of a Carnot engine to the efficiency of an Otto engine operating between the same temperatures:

11

1

O

C >−

−=

b

a

c

a

TTTT

εε

because Tc > Tb.

*79 ••• Picture the Problem We can use VP CCnR −= , VP CC=γ , and 1−γTV = a constant

to show that the entropy change for a quasi-static adiabatic expansion that proceeds from state (V1,T1) to state (V2,T2) is zero. Express the entropy change for a general process that proceeds from state 1 to state 2:

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

1

2

1

2V lnln

VVnR

TTCS


1507

For an adiabatic process: 1

2

1

1

2

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

γ

VV

TT


( )( )[ ]V

1

2

2

1

2

1V

1

2

1

2

1

2

1V

1

2

1

2

1

2

1V

1lnln

ln1ln

ln

lnlnlnln

CnRVV

VV

VVC

nRVV

VVVVC

nRVV

VVnR

VVCS

−−⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆

−

−

γγ

γ

γ

Use the relationship between CP and CV to obtain:

VP CCnR −=

Substitute for nR and γ and simplify:

0

1ln VV

pVP

1

2

=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−−⎟⎟

⎠

⎞⎜⎜⎝

⎛=∆ C

CC

CCVVS

80 ••• Picture the Problem (a) Suppose the refrigerator statement of the second law is violated in the sense that heat Qc is taken from the cold reservoir and an equal mount of heat is transferred to the hot reservoir and W = 0. The entropy change of the universe is then ∆Su = Qc/Th − Qc/Tc. Because Th > Tc, Su < 0, i.e., the entropy of the universe would decrease. (b) In this case, is heat Qh is taken from the hot reservoir and no heat is rejected to the cold reservoir, i.e., Qc = 0, then the entropy change of the universe is ∆Su = −Qh/Th + 0, which is negative. Again, the entropy of the universe would decrease. (c) The heat-engine and refrigerator statements of the second law only state that some heat must be rejected to a cold reservoir and some work must be done to transfer heat from the cold to the hot reservoir, but these statements do not specify the minimum amount of heat rejected or work that must be done. The statement ∆Su ≥ 0 is more restrictive. The heat-engine and refrigerator statements in conjunction with the Carnot efficiency are equivalent to ∆Su ≥ 0.

Chapter 19

1508

81 ••• Picture the Problem We can express the net efficiency of the two engines in terms of W1, W2, and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1, W2, Qh, and Qm. Express the net efficiency of the two engines connected in series:

h

21net Q

WW +=ε

Express the efficiencies of engines 1 and 2: h

11 Q

W=ε

and

m

22 Q

W=ε

Solve for W1 and W2 and substitute to obtain:

2h

m1

h

m2h1net εεεεε

QQ

QQQ

+=+

=

Express the efficiency of engine 1 in terms of Qm and Qh: h

m1 1

QQ

−=ε

Solve for Qm/ Qh:

1h

m 1 ε−=QQ

Substitute to obtain: ( ) 211net 1 εεεε −+=

*82 ••• Picture the Problem We can express the net efficiency of the two engines in terms of W1, W2, and Qh and then use ε1 = W1/Qh and ε2 = W2/Qm to eliminate W1, W2, Qh, and Qm. Finally, we can substitute the expressions for the efficiencies of the ideal reversible engines to obtain hcnet 1 TT−=ε .

Express the efficiencies of ideal reversible engines 1 and 2:

h

m1 1

TT

−=ε (1)

and

m

c2 1

TT

−=ε (2)

Express the net efficiency of the two engines connected in series:

h

21net Q

WW +=ε (3)


1509

Express the efficiencies of engines 1 and 2: h

11 Q

W=ε and

m

22 Q

W=ε

Solve for W1 and W2 and substitute in equation (3) to obtain:

2h

m1

h

m2h1net εεεεε

QQ

QQQ

+=+

=

Express the efficiency of engine 1 in terms of Qm and Qh: h

m1 1

QQ

−=ε

Solve for Qm/ Qh:

1h

m 1 ε−=QQ

Substitute to obtain: ( ) 211net 1 εεεε −+=

Substitute for ε1 and ε2 and simplify to obtain:

h

c

h

c

h

m

h

m

m

c

h

m

h

mnet

11

11

TT

TT

TT

TT

TT

TT

TT

−=−+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=ε

83 ••• Picture the Problem There are 26 letters and four punctuation marks (space, comma, period, and exclamation point) used in the English language, disregarding capitalization, so we have a grand total of 30 characters to choose from. This fragment is 330 characters (including spaces) long; there are then 30330 different possible arrangements of the character set to form a fragment this long. We can use this number of possible arrangements to express the probability that one monkey will write out this passage and then an estimate of a monkey’s typing speed to approximate the time required for one million monkeys to type the passage from Shakespeare. Assuming the monkeys type at random, express the probability P that one monkey will write out this passage:

330301

=P

Use the approximation 5.110100030 =≈ to obtain:

( )( )

4954953305.1 10

101

101 −===P

Assuming the monkeys can type at a rate of 1 character per second, it would take about 330 s to write a passage of length equal to the quotation from Shakespeare. Find the time T required for a million monkeys to type this particular passage by accident:

( )( )

( )

y10

s103.16y1s1030.3

1010s330

484

7491

6

495

≈

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×=

=T

Chapter 19

1510

Express the ratio of T to Russell’s estimate: 478

6

484

Russell

10y10y10==

TT

or Russell

47810 TT ≈

1511

Chapter 20 Thermal Properties and Processes Conceptual Problems *1 • Determine the Concept The glass bulb warms and expands first, before the mercury warms and expands.

2 • Determine the Concept The heating of the sheet causes the average separation of its molecules to increase. The consequence of this increased separation is that the area of the hole always increases. correct. is )(b

3 • Determine the Concept Actually, it can be hard boiled, but it does take quite a bit longer than at sea level. response.best theis )(c

4 • Determine the Concept Gases that cannot be liquefied by applying pressure at 20°C are those for which Tc < 293 K. These are He, Ar, Ne, H2, O2, NO.

*5 •• (a) With increasing altitude, P decreases; from curve OF, T of the liquid-gas interface diminishes, so the boiling temperature decreases. Likewise, from curve OH, the melting temperature increases with increasing altitude. (b) Boiling at a lower temperature means that the cooking time will have to be increased. 6 • Picture the Problem We can apply the Stefan-Boltzmann law to relate the rate at which an object radiates thermal energy to its environment. Using the Stefan-Boltzmann law, relate the power radiated by a body to its temperature:

4r ATeP σ=

where A is the surface area of the body, σ is Stefan’s constant, and e is the emissivity of the object.

Because P varies with the fourth power of T, tripling the temperature increases the rate at which it radiates by a factor of 34 and correct. is )(d

Chapter 20

1512

*7 • Determine the Concept The thermal conductivity of metal and marble is much greater than that of wood; consequently, heat transfer from the hand is more rapid.

8 • (a) True (b) True (c) False. The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. (d) False. Water contracts on heating between 0°C and 4°C. (e) True 9 • Determine the Concept Because atoms are few and far between in space, the earth can not lose heat by conduction or convection. Thermal energy is radiated through space in the form of electromagnetic waves that move at the speed of light. correct. is )(c

10 • Determine the Concept Because there is little, if any, molecule-to-molecule transportation of energy into a fireplace-heated room, the mechanisms are radiation and convection.

11 • Determine the Concept In the absence of matter to support conduction and convection, radiation is the only mechanism. 12 •• Determine the Concept Because the amount of heat lost by the house is proportional to the difference between the house temperature and that of the outside air, the rate at which the house loses heat (that must be replaced by the furnace) is greater at night when the temperature of the house is kept high than when it is allowed to cool down. 13 •• Picture the Problem The rate at which heat is conducted through a cylinder is given by

xTkAdtdQI ∆∆== // where A is the cross-sectional area of the cylinder.

Express the rate at which heat is conducted through cylinder A: x

TdkI∆∆

= 2AAA π

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